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t 






V 







STRENGTH OF MATERIALS 


A COMPREHENSIVE PRESENTATION OF SCIENTIFIC METHODS 
OF LOCATING AND DETERMINING STRESSES AND 
CALCULATING THE REQUIRED STRENGTH 
AND DIMENSIONS OF BUILDING 
MATERIALS 


BY 

EDWARD R^ MAURER, B.C.E. 

PROFESSOR OF MECHANICS, UNIVERSITY OF WISCONSIN 




ILLUSTRATED 


« •> *> 


> 




AMERICAN TECHNICAL SOCIETY 
CHICAGO 

1918 





copyright, 1907, 1914, 1917, by 
AMERICAN TECHNICAL SOCIETY 


COPYRIGHTED IN GREAT BRITAIN 
ALL RIGHTS RESERVED 


PEB -I 1918 

S t 
( < I 

©CI.A481578 

Wu \ 



INTRODUCTION 


E VERY layman is fascinated by a great engineering work—a 
large sewage system, a Keokuk dam, a twenty-story steel 
structure—and wishes he might be able to construct such work 
and carry it to completion. And yet he hardly appreciates the 
knowledge, experience, and judgment necessary to bring such a 
work to a satisfactory close. Time was when all of the details of 
such structures were determined by guesswork, but the develop¬ 
ments in science and mathematics have changed all that. For¬ 
mulas for the various types of stresses have been worked out; 
constants for every known material have been collected; and a 
multitude of diagrams and tables contribute to making the engi¬ 
neer’s work as precise as a bookkeeper’s balance. The strength 
and size of every rivet and the length and cross-section of every 
girder in a steel structure are figured so they will bear the strains 
put upon them. The design of a masonry dam, the strength of 
the concrete mixture, and the amount of steel reinforcement are 
all mathematically determined in order to safely restrain the given 
volume of water behind it. 

<1 The final judgment, therefore, as to the size of every part of 
the structure—must depend upon the designer’s knowledge of 
'‘Strength of Materials.” The treatment of Strength of Materials, 
as found in standard textbooks, is so clothed in abstruse mathe¬ 
matics that it is impossible for the average trained man to obtain 
a working knowledge of the subject. As the subject is one of the 
most important in any of the engineering branches, the author 
has thought it wise to present it in simple form, culling out all 
material that can not be used in practical design and analyzing 
the subject from the theory through to the practical formulas 
without the use of higher mathematics. In fact, he has used only 
such mathematics as may be easily understood. While the author 
has designed this work especially for home study purposes, the 
material is valuable to the college trained man as well, as it gives 
in clear, concise form the principles which are most used in engi¬ 
neering and architectural work. It is the hope of the publishers 
that the book will fill a place among the useful reference works in 
the field of engineering. 




LONGITUDINAL SECTION C-D 

SECTIONS THROUGH REINFORCED CONCRETE FLOOR, SHOWING TEST OF FLOOR 

Reproduced by Courtesy of the Expanded Metal & Corrugated Bar Co., St. Louis, Mo. 























































































CONTENTS 


Simple stress. 

Kinds of stress. 

Tension, compression, shear 

Unit-stress. 

Deformation. 

Elasticity. 

Hooke’s law, and elastic limit. 

Ultimate strength. 

Stress-deformation diagram. 

Working stress and strength. 

Factor of safety. 

Strength of materials in tension. 

Strength of materials in compression 
Strength of materials in shear. 


PAGE 
. 1 
. 2 
. 2 
. 3 
. 4 
. 4 


6 

6 

8 

8 

11 

13 

15 


Reaction of supports. 16 

Moment of a force. 16 

Principle of moments. 17 

Kinds of beams. 19 

Determination of reactions on beams. 19 

External shear and bending moment. 23 

External shear. 23 

Rule of signs. 23 

Units for shears. 24 

Shear diagrams. 27 

Maximum shear. 31 

Bending moment. 32 

Moment diagrams. 36 

Maximum bending moment. 40 

Table of maximum shears, moments, etc. 41 

Center of gravity and moment of inertia. 41 

Center of gravity of an area. 41 

Principle of moments applied to areas. 42 

Center of gravity of built-up sections. 44 

Moment of inertia. 46 

Table of centers of gravity and moments of inertia. 52 

Strength of beams. 52 

Kinds of loads. 52 

Transverse, longitudinal, inclined forces. 52 

Neutral surface. 54 

Neutral line. 54 

Neutral axis. 54 

Stress at cross-section. 55 

Fibre stress. 57 













































CONTENTS 


Strength of beams (Continued) PAGE 

Value of the resisting moment. 58 

First beam formula. 59 

Applications of the first beam formula. 59 

Laws of strength of beams. 71 

Modulus of rupture. 72 

Resisting shear. 73 

Second beam formula. 73 

Horizontal shear. 75 

Design of timber beams. 76 

Kinds of loads and beams. 79 

Flexure and tension. 79 

Flexure and compression. 81 

Combined flexural and direct stress. 83 

Strength of columns. 85 

End conditions. 85 

Classes of columns. 86 

Cross sections of columns. 86 

Radius of gyration. 86 

Kinds of column loads. 88 

Rankine’s column formula. 88 

Graphical representation of column formulas. 93 

Combination column formulas. 94 

Straight-line and Euler formulas. 94 

Parabola-Euler formulas. 98 

Broken straight-line formula.101 

Design of columns.102 

Strength of shafts.105 

Twisting moment.105 

Torsional stress.106 

Resisting moment.107 

Formula for the strength of a shaft.107 

Formula for the power which a shaft can transmit.109 

Stiffness of rods, beams, and shafts.110 

Coefficient of elasticity.Ill 

Temperature stresses.113 

Deflection of beams.114 

Twist of shafts. 116 

Non-elastic deformation.117 

Riveted joints.118 

Kinds of joints.118 

Shearing strength, or shearing value, of a rivet.118 

Bearing strength, or bearing value, of a plate.119 

Frictional strength of a joint.119 

Tensile and compressive strength of riveted plates.120 

Computation of the strength of a joint.120 

Efficiency of a joint.122 




















































Dantiic Fir ------. - --- Swedish- DDD 



DRAWING SHOWING FRACTURE OF VARIOUS WOODS UNDER BENDING STRESS 

Reproduced from “Strength and Properties of Materialsby W. G. Kirkaldy. 






























































































































































STRENGTH OF MATERIALS- 


PART I. 


SIMPLE STRESS. 

i. Stress. When forces are applied to a body they tend in a 
greater or less degree to break it. Preventing or tending to pre¬ 
vent the rupture, there arise, generally, forces between every two 
adjacent parts of the body. Thus, when a 
load is suspended by means of an iron rod, 
the rod is subjected to a downward pull at 
its lower end and to an upward pull at its 
upper end, and these two forces tend to pull 
it apart. At any cross-section of the rod 
the iron on either side ‘‘holds fast” to that on 
the other, and these forces which the parts 
of the rod exert upon each other prevent 
the tearing of the rod. For example, in Fig. 

1, let a represent the rod and its suspended 
load, 1,000 pounds; then the pull on the 
lower end equals 1,000 pounds. If we neg¬ 
lect the weight of the rod, the pull on the 
upper end is also 1,000 pounds, as shown in 
Fig. 1 (5); and the upper part A exerts 
on the lower part B an upward pull Q equal 
to 1,000 pounds, wdiile the lower part exerts 
on the upper a force P also equal to 1,000 pounds. These two 
forces, P and Q, prevent rupture of the rod at the “section” C; at 
any other section there are two forces like P and Q preventing 
rupture at that section. 

By stress at a section of a body is meant the force which the 
part of the body on either side of the section exerts on the other. 
Thus, the stress at the section C (Fig. 1) is P (or Q), and it equals 
1,000 pounds. 

a. Stresses are usually expressed (in America) in pounds, 
sometimes in tons. Thus the stress P in the preceding article is 















2 


STRENGTH OF MATERIALS 


1,000 pounds, or | ton. Notice that this value has nothing to do 
with the size of the cross-section on which the stress acts. 

3. Kinds of Stress, (a) When the forces acting on a body 
(as a rope or rod) are such that they tend to tear it, the stress at 
any cross-section is called a tension or a tensile stress. The 
stresses P and Q, of Fig. 1, are tensile stresses. Stretched ropes, 
loaded “tie rods” of roofs and bridges, etc., are under tensile stress, 
(b.) When the forces acting on a body (as a short post, brick, 

etc.) are such that they tend to 
crush it, the stress at any sec¬ 
tion at right angles to the di¬ 
rection of the crushing forces is 
called a pressure or a compres¬ 
sive stress. Fig. 2 ( a) repre¬ 
sents a loaded post, and Fig. 2 
( 1 )) the upper and lower parts. 
The upper part presses down on 
B, and the lower part presses up 
on A, as shown. P or Q is 
the compressive stress in the 
post at section C. Loaded posts, 
or struts, piers, etc., are under 
compressive stress. 

(c.) When the forces acting 
on a body (as a rivet in a bridge 
joint) are such that they tend to cut or “ shear ” it across, the stress 
at a section along wdiicli there is a tendency to cut is called a shear 
or a shearing stress. This kind of stress takes its name from the 
act of cutting with a pair of shears. In a material which is being 
cut in this way, the stresses that are being “ overcome ” are shear¬ 
ing stresses. Fig. 3 ( a ) represents a riveted joint, and Fig. 3 (&) 
two parts of the rivet. The forces applied to the joint are such 
that A tends to slide to the left, and B to the right; then B exerts 
on A a force P toward the right, and A on B a force Q toward the 
left as shown. P or Q is the shearing stress in the rivet. 

Tensions, Compressions and Shears are called simple stresses. 
'Forces may act upon a body so as to produce a combination of simple 
stresses on some section; such a combination is called a complex 


Ok 

Fig. 2. 











STRENGTH OF MATERIALS 


3 


stress. The stresses in beams are usually complex. There are other 
terms used to describe stress; they will be defined farther on. 

4. Unit=Stress. It is often necessary to specify not merely 
the amount of the entire stress which acts on an area, but also the 
amount which acts on each unit of area (square inch for example). 
By unit-stress is meant stress per unit area. 

To find the value of a unit-stress: Divide the whole stress by 
the whole area of the section on which it acts, or over which it is 
distributed. Thus, let 

P denote the value of the whole stress, 

A the area on which it acts, and 
S the value of the unit-stress; then 

S=-?~, also P = AS. (i) 

Strictly these formulas apply only when the stress P is uniform, 



^ b 

Fig. 3. 


that is, when it is uniformly distributed over the area, each square 
inch for example sustaining the same amount of stress. When 
the stress is not uniform, that is, when the stresses on different 
square inches are not equal, then P-v-A equals the average value 
of the unit-stress. 

5. Unit-stresses are usually expressed (in America) in 
pounds per square inch, sometimes in tons per square inch. If 
P and A in equation 1 are expressed in pounds and square 
inches respectively, then S will be in pounds per square inch; and 
if P and A are expressed in tons and square inches, S will be in 
tons per square inch. 

Examples. 1 . Suppose that the rod sustaining the load in 
Fig. 1 is 2 square inches in cross-section, and that the load weighs 
l f 000 pounds. What is the value of the unit-stress ? 














4 


STRENGTH OF MATERIALS 


Here P = 1,000 pounds, A= 2 square inches; hence. 

S — -L= 500 pounds per square inch. 

2. Suppose that the rod is one-half square inch in cross-sec¬ 
tion. What is the value of the unit-stress ? 

A = —y-square inch, and, as before, P= 1,000 pounds; hence 


S = 1,000-:—= 2,000 pounds per square inch. 


Notice that one must always divide the whole stress by the area to get 
the unit-stress, whether the area is greater or less than one. 

6 . Deformation. Whenever forces are applied to a body it 
changes in size, and usually in shape also. This change of size 
and shape is called deformation. Deformations are usually meas¬ 
ured in inches; thus, if a rod is stretched 2 inches, the “elonga¬ 
tion”— 2 inches. 

7 . Unit-Deformation. It is sometimes necessary to specify 
not merely the value of a total deformation but its amount per 
unit length of the deformed body. Deformation per unit length 
of the deformed body is called unit-deformation. 

To find the value of a unit-deformation: Divide the whole 
deformation by the length over which it is distributed. Thus, if 
D denotes the value of a deformation, 
l the length, 

s the unit-deformation, then 

also T)=ls. ( 2 ) 

Both D and l should always be expressed in the same unit. 

Example. Suppose that a 4-foot rod is elongated inch. 
What is the value of the unit-deformation? 

Here D=J inch, and Z=4 feet=48 inches; 
hence ^=^-^-48=^ inch per inch. 

That is, each inch \s elongated fa inch. 

Unit-elongations are sometimes expressed in per cent. To 
express a-n elongation in per cent: Divide the elongation in inches 
by the original length in inches, and multiply by 100. 

8 . Elasticity. Most solid bodies when deformed will regain 
more or less completely their natural size and shape when the de« 





STRENGTH OF MATERIALS 


5 


forming forces cease to act. This property of regaining size and 
shape is called elasticity. 

We may classify bodies into kinds depending on the degree 
of elasticity which they have, thus: 

1. Perfectly elastic bodies; these will regain their orig¬ 
inal form and size no matter how large the applied forces are if 
less than breaking values. Strictly there are no such materials, 
but rubber, practically, is perfectly elastic. 

2. Imperfectly elastic bodies; these will fully regain their 
original form and size if the applied forces are not too large, and 
practically even if the loads are large but less than the breaking 
value. Most of the constructive materials belong to this class. 

3. Inelastic or plastic bodies; these will not regain in the 
least their original form when the applied forces cease to act. Clay 
and putty are good examples of this class. 

9 . Hooke’s Law, and Elastic Limit. If a gradually increas¬ 
ing force is applied to a perfectly elastic material, the deformation 
increases proportionally to the force; that is, if P and P' denote 
two values of the force (or stress), and D and D' the values of the 
deformation produced by the force, 

then P:P':: D:D'. 

This relation is also true for imperfectly elastic materials, 
provided that the loads P and P' do not exceed a certain limit depend¬ 
ing on the material. Beyond this limit, the deformation increases 
much faster than the load; that is, if within the limit an addition 
of 1,000 pounds to the load produces a stretch of 0.01 inch, beyond 
the limit an equal addition produces a stretch larger and usually 
much larger than 0.01 inch. 

Beyond this limit of proportionality a part of the deformation 
is permanent; that is, if the load is removed the body only partially 
recovers its form and size. The permanent part of a deformation 
is called set. 

The fact that for most materials the deformation is propor¬ 
tional to the load within certain limits, is known as Hooke’s Law. 
The unit-stress within which Hooke’s law holds, or above which 
the deformation is not proportional to the load or stress, is called 
elastic limit. 


6 


STRENGTH OF MATERIALS 


10. Ultimate Strength. By ultimate tensile, compressive, 
or shearing strength of a material is meant the greatest tensile, 
compressive, or shearing unit-stress which it can withstand. 

As before mentioned, when a material is subjected to an in¬ 
creasing load the deformation increases faster than the load beyond 
the elastic limit, and much faster near the stage of rupture. Not 
only do tension bars and compression blocks elongate and shorten 
respectively, but their cross-sectional areas change also; tension 
bars thin down and compression blocks “swell out” more or less. 
The value of the ultimate strength for any material is ascertained 
by subjecting a specimen to a gradually increasing tensile, com¬ 
pressive, or shearing stress, as the case may be, until rupture oc¬ 
curs, and measuring the greatest load. The breaking load divided 
by the area of the original cross-section sustaining the stress, is the 
value of the ultimate strength. 

Example. Suppose that in a tension test of a wrought-iron 
rod inch in diameter the greatest load was 12,540 pounds. "What 
is the value of the ultimate strength of that grade of wrought iron? 

The original area of the cross-section of the rod was 
0,7854 (diameter) 2 =0.7854x 5=0.1964 square inches; hence 
the ultimate strength equals 

12,540-^0.1964=63,850 pounds per square inch, (nearly). 

11. Stress=Deformation Diagram. A “test” to determine 
the elastic limit, ultimate strength, and other information in re¬ 
gard to a material is conducted by applying a gradually increasing 
load until the specimen is broken, and noting the deformation cor¬ 
responding to many values of the load. The first and second col¬ 
umns of the following table are a record of a tension test on a steel 
rod one inch in diameter. The numbers in the first column are 
the values of the pull, or the loads, at which the elongation of 
the specimen was measured. The elongations are given in the sec¬ 
ond column. The numbers in the third and fourth columns are 
the values of the unit-stress and unit-elongation corresponding to 
the values of the load opposite to them. The numbers in the 
third column were obtained from those in the first by dividing 
the latter by the area of the cross-section of the rod, 0.7854 
square inches. Thus, 

3,930-^0.7854=5,000 
7,850-^0.7854=10,000, etc. 


STRENGTH OF MATERIALS 


7 


Total Pull 
in pounds, P 


3930 

7850 

11780 

15710 

19635 

23560 

27190 

31415 

35345 

39270 

43200 

47125 

51050 

54980 

58910 

62832 

65200 


Deformation 
in inches, D 


0.00136 

.00280 

.00404 

.00538 

.00672 

.00805 

.00942 

.01080 

.01221 

.0144 

.0800 

.1622 

.201 

.281 

.384 

.560 

1.600 


Unit-Stress in 
pounds per 
square inch, S 


5000 

10000 

15000 

20000 

25000 

30000 

35000 

40000 

45000 

50000 

55000 

60000 

65000 

70000 

75000 

80000 

83000 


Unit- 

Deformation, 

s 


0.00017 

.00035 

.00050 

.00067 

.00084 

.00101 

.00118 

.00135 

.00153 

.00180 

.0100 

.0202 

.0251 

.0351 

.048 

.070 

.200 


Ihe numbers in the fourth column were obtained by dividing 
those in the second by the length of the specimen (or rather the 
length of that part whose elongation was measured), 8 inches. 
Thus, 


0.00136-^8 = 0.00017, 

.00280-^8— .00035, etc. 

Looking at the first two columns it will be seen that the elonga¬ 
tions are practically proportional to the loads up to the ninth load, 
the increase of stretch for each increase in load beincr about 0.00135 
inch; but beyond the ninth load the increases of stretch are much 
greater. Hence the elastic limit was reached at about the ninth 
load, and its value is about 45,000 pounds per square inch. The 
greatest load was 65,200 pounds, and the corresponding unit-stress, 
83,000 pounds per square inch, is the ultimate strength. 

N early all the information revealed by such a test can be 
well represented in a diagram called a stress-deformation diagram. 
It is made as follows: Lay off the values of the unit-deformation 
(fourth column) along a horizontal line, according to some con¬ 
venient scale, from some fixed point in the line. At the points on 
the horizontal line representing the various unit-elongations, lay 
off perpendicular distances equal to the corresponding unit-stresses. 
Then connect by a smooth curve the upper ends of all those dis¬ 
tances, last distances laid off. Thus, for instance, the highest unit- 












8 


STRENGTH OF MATERIALS 


elongation (0.20) laid off from o (Fig. 4) fixes the point a , and a 
perpendicular distance to represent tlie highest unit-stress (83,000) 
fixes the point b. All the points so laid off give the curve ocb. The 
part oc, within the elastic limit, is straight and nearly vertical 
while the remainder is curved and more or less horizontal, especially 
toward the point of rupture b. Fig. 5 is a typical stress-defor¬ 
mation diagram for timber, cast iron, wrought iron, soft and hard 
steel, in tension and compression. 

12. Working Stress and Strength, and Factor of Safety. 
The greatest unit-stress in any part of a structure when it is sus¬ 
taining its loads is called the 
working stress of that part. If 
it is under tension, compression 
and shearing stresses, then the 
corresponding highest unit- 
stresses in it are called its work¬ 
ing stress in tension, in com¬ 
pression, and in shear respect¬ 
ively; that is, we speak of as 
many working stresses as it has 
kinds of stress. 

By working strength of a material to be used for a certain 
purpose is meant the highest unit-stress to which the material 
ought to be subjected when so used. Each material has a working 
strength for tension, for compression, and for shear, and they are in 
general different. 

By factor of safety is meant the ratio of the ultimate strength 
of a material to its working stress or strength. Thus, if 
S u denotes ultimate strength, 

S w denotes working stress or strength, and 
/ denotes factor of safety, then 



Fig. 4. 


/ = |i; also S w = 

b w / 


( 3 ) 


When a structure w T hich has to stand certain loads is about 
to be designed, it is necessary to select working strengths or fac¬ 
tors of safety for the materials to be used. Often the selection is 
a matter of great importance, and can be wisely performed only 
by an experienced engineer, for this is a matter where hard-and- 





STRENGTH OF MATERIALS 


9 


fast rules should not govern but rather the judgment of the expert. 
But there are certain principles to be used as guides in making a 
selection, chief among which are: 

1. The working strength should be considerably below the 
elastic limit. (Then the deformations will bs small and not per- 
manent.) 



2. The working strength should be smaller for parts of a 
structure sustaining varying loads than for those whose loads are 
steady. (Actual experiments have disclosed the fact that the 
strength of a specimen depends on the kind of load put upon it, 
and that in a general way it is less the less steady the load is.) 

3. The working strength must be taken low for non-uniform 
material, where poor workmanship may be expected, when the 





































10 


STRENGTH OE MATERIALS 


loads are uncertain, etc. Principles 1 and 2 have been reduced 
to figures or formulas for many particular cases, but the third must 
remain a subject for display of judgment, and even good guessing 
in many cases. 

The following is a table of factors of safety* which will be 
used in the problems: 


Factors of Safety. 


Materials. 

For steady 
stress. 

(Buildings.) 

For varying 
stress. 
(Bridges.) 

For shocks. 
(Machines.) 

Timber 

8 

10 

15 

Brick and stone 

15 

25 

30 

Cast iron 

6 

15 

20 

Wrought iron 

4 

6 

10 

Steel 

5 

7 

15 


They must be regarded as average values and are not to be 
adopted in every case in practice. 

Examples. 1. A wrought-iron rod 1 inch in diameter sus¬ 
tains a load of 30,000 pounds. What is its working stress? If 
its ultimate strength is 50 y 000 pounds per square inch, what is 
its factor of safety ? 

The area of the cross-section of the rod equals 0.7854 X (diam- 
eter) 2 =0.7854x 1 2 =0.7854 square inches. Since the whole stress 
on the cross-section is 30,000 pounds, equation 1 gives for the 
unit working stress 


_ 30,000 
b ~* 07854 


= 38,197 pounds per square inch. 


Equation 3 gives for factor of safety 

„ 50,000 

/= 


1.3 


38,197' 

2. How large a steel bar or rod is needed to sustain a steady 
pull of 100,000 pounds if the ultimate strength of the material is 
65,000 pounds ? 

The load being steady, we use a factor of safety of 5 (see table 
above); hence the working strength to be used (see equation 3) is 

65.0°° 

13,000 pounds per square inch. 


S 


5 


The proper area of the cross-section of the rod can now be com- 
puted from equation 1 thus: 


*Taken from Merriman’s “Mechanics of Materials.'’ 
















STRENGTH OF MATERIALS 


11 



100,000 

~ i3 000 “ ‘*692 square inches. 


A bar 2x4 inches in cross-section would be a little stronger 
than necessary. To find the diameter (^) of a round rod of suffi¬ 
cient strength, we write 0.7854 d 2 = 7.692, and solve the equation 
for d; thus: 

7.692 

d 2 = = 9.794, or <^ = 3.129 inches. 


3. IIow large a steady load can a short timber post safely sus¬ 
tain if it is 10x10 inches in cross-section and its ultimate com¬ 
pressive strength is 10,000 pounds per square inch ? 

According to the table (page 12) the proper factor of safety is 
8, and hence the working strength according to equation 3 is 


S 


10,000 
~ 8 ~ 


1,250 pounds per square inch. 


The area of the cross-section is 100 square inches; hence the safe 
load (see equation 1) is 

P = 100 X 1,250 = 125,000 pounds. 

4. When a hole is punched through a plate the shearing 
strength of the material has to be overcome. If the ultimate shear¬ 
ing strength is 50,000 pounds per square inch, the thickness of the 
plate -J inch, and the diameter of the hole | inch, what is the value 
of the force to be overcome ? 

The area shorn is that of the cylindrical surface of the hole 
or the metal punched out; that is 
3.1416 X diameter X thickness = 3.1416 X f X J = 1.178 sq. in. 
Hence, by equation 1, the total shearing strength or resistance 
to punching is 

P = 1.178 X 50,000= 58,900 pounds. 

STRENGTH OF MATERIALS UNDER SIMPLE STRESS. 


13. Materials in Tension. Practically the only materials 
used extensively under tension are timber, wrought iron and steel, 
and to some extent cast iron. 

14. Timber. A successful tension test of wood is difficult, 
as the specimen usually crushes at the ends when held in the test¬ 
ing machine, splits, or fails otherwise than as desired. Hence the 













12 


STRENGTH OF MATERIALS 


tensile strengths of woods are not well known, but the following 

may be taken as approximate average values of the ultimate 

strengths of the woods named, when “dry out of doors. ,, 

Hemlock, 7,000 pounds per square inch. 

White pine, 8,000 “ “ 

Yellow pine, long leaf, 12,000 “ “ 

“ “ , short leaf, 10,000 “ “ 

Douglas spruce, 10,000 “ “ 

White oak, 12,000 

Red oak, 9,000 “ “ 

15. Wrought Iron. The process of the manufacture of 
wrought iron gives it a “grain,” and its tensile strengths along and 
across the grain are unequal, the latter being about three-fourths 
of the former. The ultimate tensile strength of wrought iron 
along the grain varies from 45,000 to 55,000 pounds per square 
inch. Strength along the grain is meant when not otherwise 
stated. 

The strength depends on the size of the piece, it being greater 
for small than for large rods or bars, and also for thin than for 
thick plates. The elastic limit varies from 25,000 to 40,000 
pounds per square inch, depending on the size of the bar or plate 
even more than the ultimate strength. Wrought iron is very 
ductile, a specimen tested in tension to destruction elongating from 
5 to 25 per cent of its length. 

16. Steel. Steel has more or less of a grain but is practically 
of the same strength in all directions. To suit different purposes, 
steel is made of various grades, chief among which may be men¬ 
tioned rivet steel, sheet steel (for boilers), medium steel (for 
bridges and buildings), rail steel, tool and spring steel. In general, 
these grades of steel are hard and strong in the order named, the 
ultimate tensile strength ranging from about 50,000 to 160,000 
pounds per square inch. 

There are several grades of structural steel, which may be 
described as follows:* 

1. Rivet steel: 

Ultimate tensile strength, 48,000 to 58,000 pounds per square inch. 

Elastic limit, not less than one-half the ultimate strength. 

Elongation, 26 per cent. 

Bends 180 degrees flat on itself without fracture. 


*Takenfrom “Manufacturer’s Standard Specifications.” 







STRENGTH OF MATERIALS 


13 


2. Sort steel: 

Ultimate tensile strength, 52,000 to G2,000 pounds per square inch. 

Elastic limit, not less than one-half the ultimate strength. 

Elongation, 25 per cent. 

Bends 180 degrees flat on itself. 

3. Medium steel: 

Ultimate tensile strength, 60,000 to 70,000 pounds per square inch. 

Elastic limit, not less than one-half the ultimate strength. 

Elongation, 22 per c®nt. 

Bends 180 degrees to a diameter equal to the thickness of the 
specimen without fracture. 

17. Cast Iron. As in the case of steel, there are many 
grades of cast iron. The grades are not the same for all localities 
or districts, hut they are based on the appearance of the fractures, 
which vary from coarse dark grey to fine silvery white. 

The ultimate tensile strength does not vary uniformly with 
the grades but depends for the most part on the percentage of 
“combined carbon” present in the iron. This strength varies from 
15,000 to 35,000 pounds per square inch, 20,000 being a fair 
average. 

Cast iron has no well-defined elastic limit (see curve for cast 
iron, Fig. 5). Its ultimate elongation is about one per cent. 

EXAMPLES FOR PRACTICE. 

1. A steel wire is one-eightli inch in diameter, and the ulti¬ 
mate tensile strength of the material is 150,000 pounds per square 
inch. IIow large is its breaking load ? Ans. 1,845 pounds. 

2. A wrought-iron rod (ultimate tensile strength 50,000 

pounds per square inch) is 2 inches in diameter. How large a 
steady pull can it safely hear ? Ans. 39,270 pounds. 

18. Materials in Compression. Unlike the tensile, the 
compressive strength of a specimen or structural part depends on 
its dimension in the direction in which the load is applied, for, 
in compression, a long bar or rod is weaker than a short one. At 
present we refer only to the strength of short pieces such as do 
not bend under the load, the longer ones (columns) being dis¬ 
cussed farther on. 

Different materials break or fail under compression, in two 
very different ways: 

1. Ductile materials (structural steel, wrought iron, etc.), 


14 


STRENGTH OF MATERIALS 


and wood compressed across the grain, do not fail by breaking into 
two distinct parts as in tension, but tlie former bulge out and 
flatten under great loads, while wood splits and mashes down. 
There is no particular point or instant of failure under increasing 
loads, and such materials have no definite ultimate strength in 
compression. 

2. Brittle materials (brick, stone, hard steel, cast iron, etc.), 
and wood compressed along the grain, do not mash gradually, but 
fail suddenly and have a definite ultimate strength in compression. 
Although the surfaces of fracture are always much inclined to the 
direction in which the load is applied (about 45 degrees), the ulti¬ 
mate strength is computed by dividing the total breaking load by 
the cross-sectional area of the specimen. 

The principal materials used under compression in structural 
work are timber, wrought iron, steel, cast iron, brick and stone. 

19. Timber. As before noted, timber has no definite ulti¬ 
mate compressive strength across the grain. The U. S. Forestry 
Division has adopted certain amounts of compressive deformation 
as marking stages of failure. Three per cent compression is 
regarded as “a working limit allowable,” and fifteen per cent as 
“an extreme limit, or as failure.” The following (except the first) 
are values for compressive strength from the Forestry Division 
Reports, all in pounds per square inch: 


Hemlock. 

Ultimate strength 
along the grain. 

6,000 

3 % Compression 
across the grain 

White pine. 

5,400 

700 

Long-leaf yellow pine.. 

8,000 

1,260 

Short-leaf yellow pine.. 

6,500 

1,050 

Douglas spruce. 

5,700 

800 

White oak. 

8,500 

2,200 

Red oak. 

7,200 

2,300 


20. Wrought Iron. The elastic limit of wrought iron, as be¬ 
fore noted, depends very much upon the size of the bars or plate, it 
being greater for small bars and thin plates. Its value for com¬ 
pression is practically the same as for tension, 25,000 to 40,000 
pounds per square inch. 

21. Steel. The hard steels have the highest compressive 
strength; there is a recorded value of nearly 400,000 pounds per 
square inch, but 150,000 is probably a fair average. 









STRENGTH OF MATERIALS 


15 


Tlie elastic limit in compression is practically the same as in 
tension, which is about 60 per cent of the ultimate tensile strength, 
or, for structural steel, about 25,000 to 42,000 pounds per square 
inch. 

22. Cast Iron. This is a very strong material in compres¬ 
sion, in which way, principally, it is used structurally. Its ulti¬ 
mate strength depends much on the proportion of “combined car¬ 
bon” and silicon present, and varies from 50,000 to 200,000 pounds 
per square inch, 90,000 being a fair average. As in tension, 
there is no well-defined elastic limit in compression (see curve for 
cast iron, Fig. 5). 

2 3 . Brick. The ultimate strengths are as various as the 
kinds and makes of brick. For soft brick, the ultimate strength 
is as low as 500 pounds per square inch, and for pressed brick it 
varies from 4,000 to 20,000 pounds per square inch, 8,000 to 
10,000 being a fair average. The ultimate strength of good pav¬ 
ing brick is still higher, its average value being from 12,000 to 
15,000 pounds per square inch. 

24. Stone. Sandstone, limestone and granite are the 
principal building stones. Their ultimate strengths in pounds 
per square inch are about as follows: 

Sandstone,* 5,000 to 16,000, average 8,000. 

Limestone,* 8,000 “ 16,000, “ 10,000. 

Granite, 14,000 “ 24,000, « 16,000. 

Compression at right angles to the “bed” of the stone. 

EXAMPLES FOR PRACTICE. 

1. A limestone 12x12 inches on its bed is used as a pier 

cap, and bears a load of 120,000 pounds. What is its factor of 
safety ? Ans. 12. 

2. How large a post (short) is needed to sustain a steady 
load of 100,000 pounds if the ultimate compressive strength of 
the wood is 10,000 pounds per square inch ? Ans. 8 X10 inches. 

25. Materials in Shear. The principal materials used under 
shearing stress are timber, wrought iron, steel and cast iron. 
Partly on account of the difficulty of determining shearing 
strengths, these are not well known. 

26. Timber. The ultimate shearing strengths of the more 
important woods along the gxain are about as follows: 


16 


STRENGTH OF MATERIALS 


Hemlock, 

White pine, 

Long-leaf yellow pine, 
Short-leaf “ “ 

Douglas spruce, 

White oak, 

Red oak, 


300 pounds per square inch. 
400 “ “ 

850 “ “ 

775 “ “ 

500 

1,000 “ “ 

1,100 “ “ 


Wood rarely fails by shearing across the grain 


Its ultimate 


C 


Fig. 6 a. Fig. 6 b. 

shearing strength in that direction is probably four or five times 
the values above given. 

27. Metals. The ultimate shearing strength of wrought 
iron, steel, and cast iron is about 80 per cent of their respective 
ultimate tensile strengths. 




EXAMPLES FOR PRACTICE. 

1. How large a pressure P (Pig. 6 a ) exerted on the shaded 
area can the timber stand before it will shear off on the surface 
abed , if ah — 6 inches and be = 10 inches, and the ultimate shear- 
ing strength of the timber is 400 pounds per square inch ? 

Ans. 24,000 pounds. 

2. When a bolt is under tension, there is a tendency to tear 
the bolt and to “strip” or shear off the head. The shorn area 
would be the surface of the cylindrical hole left in the head. 
Compute the tensile and shearing unit-stresses when P (Fig. 6 
equals 30,000 pounds, d ~ 2 inches, and t — 3 inches. 

^ ( Tensile unit-stress, 9,550 pounds per square inch. 

* ( Shearing unit-stress, 1,591 pounds per square inch. 

REACTIONS OF SUPPORTS. 


28 . Moment of a Force. By moment of a force with re¬ 
spect to a point is meant its tendency to produce rotation about 
that point. Evidently the tendency depends on the magnitude of 
the force and on the perpendicular distance of the line of action 
of the force from the point : the greater the force and the per¬ 
pendicular distance, the greater the tendency; hence the moment 



















STRENGTH OF MATERIALS 


17 


of a force with respect to a point equals the product of the force 
and the perpendicular distance from the force to the point. 

The point with respect to which the moment of one or more 
forces is taken is called an origin or center of moments, and the 
perpendicular distance from an origin of moments to the line of 
action of a force is called the arm of the force with respect to 
that origin. Thus, if and F 2 (Fig. 7) are forces, their arms 
with respect to O' are af and a 2 ' respectively, and their moments 
are F x a\ and ~F 2 a' 2 . With respect to 0" their arms are af and af 
respectively, and their moments are F^/' and F 2 af. 

If the force is expressed in pounds and its arm in feet, the 
moment is in foot-pounds; if the force is in pounds and the arm 
in inches, the moment is in inch-pounds. 

29. A sign is given to the moment of a force for conven¬ 
ience; the rule used herein is as follows: The moment of a 
force about a point is positive or negative according as it tends 
to turn the body about that p>oint in the clockwise or counter¬ 
clockwise direct ion *. 


Thus the moment (Fig. 7) 

of Fj about O' is negative, about O" positive; 


u TT a O' “ a 

x 2 

30. Principle of Moments. 

proper magnitude and line of ac¬ 
tion can balance any number of 
forces. That single force is called 
the equilibrant of the forces, and 
the single force that would balance 
the equilibrant is called the result¬ 
ant of the forces. Or, otherwise 
stated, the resultant of any num¬ 
ber of forces is a force which pro¬ 
duces the same effect. It can be 
proved that— The algebraic sum 
of the moments of any number 
of forces with respect to a point , 
equals the moment of their re¬ 
sultant about that point. 


, about O" negative. 

' O 

I 11 general, a single force of 



*By clockwis© direction is meant that in which the hands of a clock 
rotate; and by counter-clockwise, the opposite direction. 






18 


STRENGTH OF MATERIALS 


This is a useful principle and is called “principle of moments.” 
31. All the forces acting upon a body which is at rest are 

said to be balanced or in equilibrium. No force is required to 
balance such forces and hence their equilibrant and resultant are 
zero. 

Since their resultant is zero, the algebraic sum of the mom- 

looollos. 2000II0S. 2000II0S. loooltos. 








> 

4 c 1 

£__ 

c J 
1 > 

IE ? 

P - E. 


f 

A 

k 1 

3s. 

T * 

D 


30oolbs. 3ooolbs. 

Fig. 8. 


ents of any number of forces which are balanced or in equilib¬ 
rium equals zero . 

This is known as the principle of moments for forces in 
equilibrium; for brevity we shall call it also “the principle of 
moments.” 

The principle is easily verified in a simple case. Thus, let 
AB (Fig. 8) be a beam resting on supports at C and F. It is 
evident from the symmetry of the loading that each reaction 
equals one-half of the whole load, that is, -J of 6,000=3,000 
pounds. (We neglect the weight of the beam for simplicity.) 

With respect to C, for example, the moments of the forces 
are, taking them in order from the left: 

—1,000 X 4 = — 4,000 foot-pounds 
3,000 X 0= 0 

2,000 X 2 = 4,000 

2,000 X 14 = 28,000 

—3,000 X 16 = — 48,000 
1,000 X 20 = 20,000 

The algebraic sum of these moments is seen to equal zero. 

Again, with respect to B the moments are: 

—1,000 X 24 = — 24,000 foot-pounds 
3,000 X 20 = 60,000 

—2,000 X 18 =—36,000 
— 2,000 X 6 = — 12,000 
3,000 X 4= 12,000 

1,000 X 0 = 0 

The sum of these moments also equals zero. In fact, no matter 














19 


STRENGTH OF MATERIALS 


where the center of moments is taken, it will be found in this and 
any other balanced system of forces that the algebraic sum of their 
moments equals zero. The chief use that we shall make of this 
principle is in finding the supporting forces of loaded beams. 

32. Kinds of Beams. A cantilever beam is one resting on 
one support or fixed at one end, as in a wall, the other end being 
free. 

A simple beam is one resting on two supports. 

A restrained beam is one fixed at both ends; a beam fixed at 
one end and resting on a support at the other is said to be re¬ 
strained at the fixed end and simply supported at the other. 

A continuous beam is one resting on more than two supports. 

33. Determination of Reactions on Beams. The forces which 
the supports exert on a beam, that is, the “supporting forces,” are 
called reactions. We shall deal chiefly with simple beams. The 
reaction on a cantilever beam supported at one point evidently 
equals the total load on the beam. 

When the loads on a horizontal beam are all vertical (and 


1000 lbs. aooo lbs. 3000 ]bs. 



1/ A v 

r C 1 5 


c *»• ■J 

I* 1 - y 

> 

t > 

* — c. 

t 

J 

" B CD 


Fig. 9. 


this is the usual case), the supporting forces are also vertical and 
the sum of the reactions equals the sum of the loads. This prin¬ 
ciple is sometimes useful in determining reactions, but in the case 
of simple beams the principle of moments is sufficient. The gen¬ 
eral method of determining reactions is as follows: 

1. Write out two equations of moments for all the forces 
(loads and reactions) acting on the beam with origins of moments 
at the supports. 

2. Solve the equations for the reactions. 

3. As a check, try if the sum of the reactions equals the 
sum of the loads. 

Examples. 1. Fig. 9 represents a beam supported at its 
ends and sustaining three loads. We wish to find the reactions 
due to these loads. 












20 


STRENGTH OF MATERIALS 


Let the reactions be denoted by Rj and R 2 as shown; then 
the moment equations are: 

For origin at A, 

1,000 X1 + 2,000 x 6 + 3,000 X 8 — R 2 X10 = 0. 

For origin at E, 


aioolbs. 36oo llos. isoolbs. 



Fig. 10. 

R 1 x 10—1,000 X 9—2,000 X 4—3,000 X 2 = 0. 

The first equation reduces to 

10 R 2 = 1,000+12,000 + 24,000 = 37,000; or 
R 2 = 3,700 pounds. 

The second equation reduces to 

10 Rj= 9,000 + 8,000 + 6,000 = 23,000; or 
R x = 2,300 pounds. 

The sum of the loads is 6,000 pounds and the sum of the reactions 
is the same; hence the computation is correct. 

2. Fig. 10 represents a beam supported at B and D (that is, 
it has overhanging ends) and sustaining three loads as shown. We 
wish to determine the reactions due to the loads. 

Let Rj and R 2 denote the reactions as shown; then the moment 
equations are: 

For origin at B, 

-2,100x2 + 0 + 3,600x6—R 2 Xl4+l,600X18 = 0. 

For origin at D, 

- 2,100 X16 ■+ R 1 X 14—3,600 X 8 + 0 + 1,600 X 4 = 0. 

The first equation reduces to 

14 R 2 — -4,200 + 21,600 + 28,800 = 46,200; or 
R 2 — 3,300 pounds. 

The second equation reduces to 

14 R 1= = 33,600 + 28,800-6,400 = 56,000; or 
R 1= 4,000 pounds. 

The sum of the loads equals 7,300 pounds and the sum of the 
reactions is the same; hence the computation checks. 

3. What are the total reactions in example 1 if the beam 
weighs 400 pounds ? 
















STRENGTH OF MATERIALS 


21 


(1.) Since we already know the reactions due to the loads 
(2,300 and 3,700 pounds at the left and right ends respectively 
(see illustration 1 above), we need only to compute the reactions 
due to the w T eight of the beam and add. Evidently the reactions 
due to the weight equal 200 pounds each; hence the 

left reaction =2,300 + 200=2,500 pounds, and the 
right “ = 3,700 + 200=3,900 “ . 

(2.) Or, we might compute the reactions due to the loads 
and weight of the beam together and directly. In figuring the 
moment due to the weight of the beam, we imagine the weight 
as concentrated at the middle of the beam; then its moments with 
respect to the left and right supports are (400 X 5) and—(400X 5) 
respectively. The moment equations for origins at A and E are 
like those of illustration 1 except that they contain one more 
term, the moment due to the weight; thus they are respectively: 

1,000 X1 + 2,000 X 6 + 3,000 X 8—R 2 X10 ■+ 400 X 5=0, 

Rj X 10—1,000 X 9—2,000 X 4—3,000 X 2—400 X 5=0. 

The first one reduces to 

10 R 2 = 39,000, or R„ = 3,900 pounds; 
and the second to 

10 Rj= 25,000, or R^ 2,500 pounds. 

4. What are the total reactions in example 2 if the beam 
weighs 42 pounds per foot ? 

As in example 3, we might compute the reactions due to the 
weight and then add them to the corresponding reactions due to 
the loads (already found in example 2), but we shall determine 
the total reactions due to load and weight directly. 

The beam being 20 feet long, its weight is 42 X 20, or 840 
pounds. Since the middle of the beam is 8 feet from the left and 
6 feet from the right support, the moments of the weight with 
to the left and right supports are respectively: 

840x8 = 6,720, and—840x6 = —5,040 foot-pounds. 

The moment equations for all the forces applied to the beam 
for origins at 13 and I) are like those in example 2, with an addi¬ 
tional term, the moment of the weight; they are respectively: 
—2,100 X 2 + 0+3,600 X 6—R 2 X 14+1,600 X 18 + 6,720 = 0, 
—2,100 X16 + Rj X H—3,600 X 8 + 0+1,600 X 4—5,040 = 0. 




22 


STRENGTH OE MATERIALS 


The first equation reduces to 

14 R 2 =52,920, or R 2 =3,780 pounds, 

and the second to 

14 R l= = 61,040, or Rj= 4,360 pounds. 

The sum of the loads and weight of beam is 8,140 pounds; 
and since the sum of the reactions is the same, the computation 
checks. 

EXAMPLES FOR PRACTICE. 

1. AB (Fig. 11) represents a simple beam supported at its 
ends. Compute the reactions, neglecting the weight of the beam. 

. ^ Right reaction = 1,443.75 pounds. 

1 ns * j Left reaction = 1,556.25 pounds. 


6 oo lbs. 900lbs. soolbs. looolbs. 




< I 1 'I 


/ N 

r 2 > 

' *+ > 

l _5 

L > 

K K> ' 

L _3 

( 

-i 


Fig. 11. 


2. Solve example 1 taking into account the weight of the 
beam, which suppose to be 400 pounds. 

. \ Right reaction = 1,643.75 pounds. 

ns * ) Left reaction = 1,756.25 pounds. 

3. Fig. 12 represents a simple beam weighing 800 pounds 
supported at A and B, and sustaining three loads as shown. 
Wh at are the reactions ? 

. ( Right reaction = 2,014.28 pounds. 

" 1Ls " ) Left reaction == 4,785.72 pounds. 


20 

L A . 

oo lbs. ioc 

olbs. 3000 

. o' . 1 

lbs. 

V _ o'_J 

K l -s 
1 > 

t 1 1 ■■■ 1 ■ 5 .. )]<—2 —» 

/ i > 



A 



Fig. 12. 


4. Suppose that in example 3 the beam also sustains a uni¬ 
formly distributed load (as a floor) over its entire length, of 500 
pounds per foot. Compute the reactions due to all the loads and 
the weight of the beam. 


Ans. 


Right reaction = 4,871.43 pounds. 
Left reaction = 11,928.57 pounds. 




























STRENGTH OF MATERIALS 


23 


EXTERNAL SHEAR AND BENDING MOMENT. 

On almost every cross-section of a loaded beam there are 
three kinds of stress, namely tension, compression and shear. The 
first two are often called fibre stresses because they act along the 
real fibres of a wooden beam or the imaginary ones of which we 
may suppose iron and steel beams composed. Before taking up 
the subject of these stresses in beams it is desirable to study certain 
quantities relating to the loads, and on which the stresses in a 
beam depend. These quantities are called external shear and 
bending moment , and will now be discussed. 

34. External Shear. By external shear at (or for) any sec¬ 
tion of a loaded beam is meant the algebraic sum of all the loads 
(including weight of beam) and reactions on either side of the 
section. This sum is called external shear because, as is shown 
later, it equals the shearing stress (internal) at the section. For 
brevity, we shall often say simply “shear” when external shear is 
meant. 

35. Rule of Signs. In computing external shears, it is cus¬ 
tomary to give the plus sign to the reactions and the minus sign 
to the loads. But in order to get the same sign for the external 
shear whether computed from the right or left, we change the sign 
of the sum when computed from the loads and reactions to the 
right. Thus for section a of the beam in Fig. 8 the algebraic sum is, 
when computed from the left, 

-1,000 + 3,000= +2,000 pounds; 
and when computed from the right, 

-1,000 + 3,000-2,000-2,000 = -2,000 pounds. 

The external shear at section a is +2,000 pounds. 

Again, for section b the algebraic sum is, 
when computed from the left, 

-1,000 + 3,000-2,000-2,000 + 3,000 = + 1,000 pounds; 
and when computed from the right, -1,000 pounds. 

The external shear at the section is +1,000 pounds. 

It is usually convenient to compute the shear at a section 
from the forces to the right or left according as there are fewer 
forces (loads and reactions) 011 the right or left sides of the 
section. 



24 


STRENGTH OF MATERIALS 


36 . Units for Shears. It is customary to express external 
shears in pounds, but any other unit for expressing force and 
weight (as the ton) may be used. 

37 . Notation. We shall use Y to stand for external shear at 
any section, and the shear at a particular section will be denoted 
by that letter subscripted; thus Y 1? Y 2 , etc., stand for the shears 
at sections one, two, etc., feet from the left end of a beam. 

The shear has different values just to the left and right of a 
support or concentrated load. We shall denote such values by Y' 
and Y"; thus Y/ and Y " denote the values of the shear at sec- 
tions a little less and a little more than 5 feet from the left end 
respectively. 

Examples. 1. Compute the shears for sections one foot 
apart in the beam represented in Fig. 9, neglecting the weight of 
the beam. (The right and left reactions are 3,700 and 2,300 
pounds respectively; see example 1, Art. 33.) 

All the following values of the shear are computed from the 
left. The shear just to the right of the left support is denoted by 
Y 0 ", and Y 0 " = 2,300 pounds. The shear just to the left of B is 
denoted by Y/, and since the only force to the left of the section 
is the left reaction, Y/= 2,300 pounds. The shear just to the 
right of B is denoted by Y/', and since the only forces to the left 
of this section are the left reaction and the 1 , 000 -pound load, 
Y/' = 2,300 —1,000= 1,300 pounds. To the left of all sections 
between B and C, there are but two forces, the left reaction and 
the 1 , 000 -pound load; hence the shear at any of those sections 
equals 2,300 - 1,000 = 1,300 pounds, or 

Y 2 = Y 3 = Y 4 = Y 5 =■ Y 6 '= 1,300 pounds. 

The shear just to the right of C is denoted by Y 0 "; and since the 
forces to the left of that section are the left reaction and the 
1 , 000 - and 2 , 000 -pound loads, 

Y 6 " = 2,300 —1,000 - 2,00t — — 700 pounds. 

Without further explanation, the student should understand 

that 

Y 7 = +2,300 - 1,000 - 2,000 = - 700 pounds, 
y; — _ 700, 

Y/' — + 2,300 - 1,000 - 2,000 - 3,000 = - 3,700, 

Y 9 = Y 10 '=-3,700, 

Y 10 " = + 2.300 - 1,000 - 2.000 - 3,000 + 3,700 = 0 















STRENGTH OF MATERIALS 


25 


2. A simple beam 10 feet long, and supported at each end, 
weighs 400 pounds, and bears a uniformly distributed load of 
1,600 pounds. Compute the shears for sections two feet apart. 

Evidently each reaction equals one-half the sum of the load 
and weight of the beam, that is, \ (1,600 + 400) =1,000 pounds. 
To the left of a section 2 feet from the left end, the forces acting 
on the beam consist of the left reaction, the load on that part of 
the beam, and the weight of that part ; then since the load and 
weight of the beam per foot equal 200 pounds, 

Y 2 = 1,000-200 X 2 = 600 pounds. 

To the left of a section four feet from the left end, the forces 
are the left reaction, the load on that part of the beam, and the 
weight ; hence 

Y 4 = 1,000-200 X 4 = 200 pounds. 

Without further explanation, the student should see that 

V 6 = 1,000-200 X 6 =-200 pounds, 

Y 8 = 1,000-200 X 8 = —600 pounds, 

V 10 ' = 1 , 000-200 X 10 = - 1,000 pounds, 

V 10 "= 1,000-200 X 10+1,000 = 0. 

3. Compute the values of the shear in example 1, taking 
into account the weight of the beam (400 pounds). (The right 
and left reactions are then 3,900 and 2,500 pounds respectively; 
see example 3, Art. 33.) 

We proceed just as in example 1, except that in each compu¬ 
tation w r e include the weight of the beam to the left of the section 
(or to the right w T hen computing from forces to the right). The 
weight of the beam being 40 pounds per foot, then (computing 
from the left) 

Y 0 " =+2,500 pounds, 

Y; =+2,500-40 =+2,460, 

Y" =+2,500-40-1,000= +1,460, 

Y 2 =+2,500-1,000-40x2= +1,420, 

Y 3 = + 2,500-1,000-40x3 =+1,380, 

Y 4 =+ 2,500-1,000-40 X 4 = +1,340, 

Y 5 =+2,500-1,000-40 X 5 = +1,300, 

Y; =+ 2,500-1,000-40 X 6 = +1,260, 

Y 6 " =+ 2,500-1,000-40 X 6-2,000 = -740, 

Y 7 =+2,500-1,000-2,000-40 X 7 = -780, 

























20 


STRENGTH OF MATERIALS 


y; = + 2,500-1,000-2,000-40 x8 =-820, 

Y 8 " = + 2,500-1,000-2,000-40 X 8-3,000 = -3,820, 

V 9 = + 2,500-1,000-2,000-3,000-40 X 9 =-3,860, 

V' 10 = +2,500-1,000-2,000-3,000-40 X10 = -3,900, 
V" 10 = + 2,500-1,000-2,000-3,000-40 X10 + 3,900=0. 

Computing from the right, we find, as before, that 

V 7 =-( 3,900-3,000-40 x3)=-780 pounds, 

V 8 ' =-(3,900-3,000-40x2) =-820, 

Y 8 " =-(3,900-40 x2)=-3,820, 

etc., etc. 

EXAMPLES FOR PRACTICE. 


1. Compute the values of the shear for sections of the beam 
represented in Fig. 10, neglecting the weight of the beam. (The 
right and left reactions are 3,300 and 4,000 pounds respectively; 
see example 2, Art. 33.) 


r Y, =Y/=-2,100 pounds, 

Ans J V" =X=V =V s =V e =V ; =V'= +1,900, 

' I v," =V 9 =V IO =Y 11 =Y 12 =Y, 3 =y u =y, 5 =Y' 16 =-l,700, 
L V", 6 =V n =V 18 =V 19 =V' 20 =+1,600. 


2. Solve the preceding example, taking into account the 
weight of the beam, 42 pounds per foot. (The right and left 
reactions are 3,780 and 4,360 pounds respectively; see example 4, 
Art. 33.) 


Ans. 


Y 0 " = - 2,100 lbs. Y 7 = + 1,966 lbs. V 14 =-1,928 lbs. 


V, 

= - 2,142 

V ' 

8 

= + 1,924 

v„ = 

- - 1,970 

V' 

= - 2,184 

V" 

= - 1,676 


- - 2,012 

V 2 " 

= + 2,176 

v. 

= -1,718 

Ve"= 

= + 1,768 

V 

= + 2,134 

Vo 

= - 1,760 

v„ = 

= + 1,726 

V, 

= + 2,092 

V„ 

= -1,802 

V. = 

= + 1,684 

V 

= + 2,050 

V„ 

= - 1,844 

V,. = 

= + 1,642 

v 6 

= + 2,008 

V,3 

= -1,886 

V ' = 
20 

= + 1,600 

Compute the 

values of 

the shear at 

sections 

one foot a 


in the beam of Fig. 11, neglecting the weight. (The right and 
left reactions are 1,444 and 1,556 pounds respectively; see example 
1, Art. 33.) 




STRENGTH OF MATERIALS 


27 


Ans. 


r~ 




v 0 " =Y 1 =Y 2 '= +1,556 pounds, 

v 2 " =v,=v 4 =v,=v;=+956, 

V 6 " =V/= + 56, 


V 7 " =V B =Y 9 =Y 10 =Y 11 =Y ia =y i8 '=_444 > 
^ 13 = Vi4=Y 15 =Y 16 '=-1,444. 


4. Compute the vertical shear at sections one foot apart in 
the beam of 4ig. 12, taking into account the weight of the beam. 
800 pounds, and a distributed load of 500 pounds per foot. (The 
right and left reactions are 4,870 and 11,930 pounds respectively; 
see examples 3 and 4, Art. 33.) 


Ans. 


VY = 

vr= 

y 2 - 

Vs = 


0 V 7 = + 6,150 lbs. 
540 lbs. V 8 ' = +5,610 
Y 8 '' =+ 4,610 
Y 9 =+4,070 
Y 10 = + 3,530 


2,540 

3,080 

3,620 


Y 4 = - 4,160 
Y 5 = - 4,700 
y; = - 5,240 
Y 6 " = +6,690 


Y n =+2,990 
V ia = +2,450 
Y u = +1,910 
V M = + 1,370 


Y 15 = + 830 lbs 
Y 16 = + 290 
Y„' = - 250 
V 17 " = -3,250 
Y I8 = - 3,790 
Y 19 = - 4,330 
y 2 ; = - 4,870 

v 20 "= 0 


38. Shear Diagrams. The way in which the external shear 
varies from section to section in a beam can be well represented 
by means of a diagram called a shear diagram . To construct 
such a diagram for any loaded beam, 

1. Lay off a line equal (by some scale) to the length of 
the beam, and mark the positions of the supports and the loads. 
(This is called a “base-lined’) 

2. Draw a line such that the distance of any point of it 
from the base equals (by some scale) the shear at the correspond¬ 
ing section of the beam, and so that the line is above the base 
where the shear is positive, and below it where negative. (This is 
called a shear line , and the distance from a point of it to the 
base is called the “ordinate” from the base to the shear line at 
that point.) 

We shall explain these diagrams further by means of illus¬ 
trative examples. 

Examples. 1. It is required to construct the shear diagram 
for the beam represented in Fig. 13, & (a copy of Fig. 9). 






28 


STRENGTH OF MATERIALS 


Lay off A'E' (Fig. 13, b) to represent the beam, and mark the 
positions of the loads B', C' and D'. In example 1, Art. 37, we 
computed the values of the shear at sections one foot apart*, hence 
we lay off ordinates at points on A'E' one foot apart, to represent 
those shears. 

Use a scale of 4,000 pounds to one inch. Since the shear for 
any section in AB is 2,300 pounds, we draw a line ccb parallel 
to the base0.575 inch(2,300-^4,000) therefrom; this is the shear 
line for the portion AB. Since the shear for any section in BO 
equals 1,300 pounds, we draw a line b'c parallel to the base and 


looollos. 


2000 lbs. 3000 lbs. 


L i 1 -J 


< ° 1 'i 

, ol J 

1 

f \ 

v - ^ ; 

/ \ 

f 1 

t 1 

1 

B 1 

c 

D 


a. 1 


b 

b’ 


I 
l 


E* 


A B‘ 


c‘ 


Scade: i"= 4-ooolbs. 


d' 


Fig. 13. 


0.325 inch (1,300-^4,000) therefrom; this is the shear line for the 
portion BC. Since the shear for any section in CD is -700 
pounds, we draw a line cd below the base and 0.175 inch 
(700-^-4,000) therefrom; this is the shear line for the portion 
CD. Since the shear for any section in DE equals -3,700 lbs., we 
draw a line^'<? below the base and 0.925 inch (3,700-^-4,000) there¬ 
from; this is the shear line for the portion DE. Fig. 13, b , is the 
required shear diagram. 

2. It is required to construct the shear diagram for the 
beam of Fig. 14, a (a copy of Fig. 9), taking into account the 
weight of the beam, 400 pounds. 

The values of the shear for sections one foot apart were com¬ 
puted in example 3, Art. 37, so we have only to erect ordinates at 
the various points on a base line A'E' (Fig. 14, &), equal to those 




























































































































STRENGTH OF MATERIALS 


29 


values. We sliall use tlie same scale as in the preceding illustra¬ 
tion, 4,000 pounds to an inch. Then the lengths of the ordinates 
corresponding to the values of the shear (see example 3, Art. 37) 
are respectively: 

2,500-^4,000=0.625 inch 

2.460- s-4,000=0.615 “ 

1.460- s-4,000=0.365 “ 

etc. etc. 

Laying these ordinates off from the base (upwards or downwards 
according as they correspond to positive or negative shears), we 
get ab , Vc, cd , and d'e as the shear lines. 


looolfos. 2oooltos. 3000 ltos. 



3. It is required to construct the shear diagram for the 
cantilever beam represented in Fig. 15, a , neglecting the weight 
of the beam. 

The value of the shear for any section in AB is - 500 pounds; 
for any section in BC, -1,500 pounds; and for any section in 
CD, - 3,500 pounds. Hence the shear lines are cd), Vc , dd. The 
scale being 5,000 pounds to an inch, 

A 'a = 500^-5,000 = 0.1 inch, 

B 'b' = 1,500-5-5,000 = 0.3 « 

C V = 3,500-f-5,000 = 0.7 “ 

The shear lines are all below the base because all the values of the 
shear are negative. 



























































































































30 


STRENGTH OF MATERIALS 


4. Suppose that the cantilever of tlie preceding illustration 
sustains also a uniform load of 200 pounds per foot (see Fig. 16, a). 
Construct a shear diagram. 


Soolbs. looolbs. '2ooo lbs. 
A 


Ai 


( O* N 


, SI' > 

W X 

/ X 


1 

! 

B 

C D 


a- 


c* 

Sc&Te i"= 5000 lbs. 

Fig. 15. 


First, we compute the values of the shear at several sections. 
Thus V 0 " —- 500 pounds, 

V, =-500- 200=- 700, 

V; =_500 - 200x2=-900, 

V 2 " =- 500 - 200x2 - 1,000=-1,900, 
y 3 =- 500 - 1,000 - 200 X 3=-2,100, 
y 4 =_ 500 - 1,000 - 200x4=-2,300, 
y r ; =-500 - 1,000 - 200x5=-2,500, 
y r ; =- 500 - 1,000 - 200x5 - 2,000=-4,500, 
y 6 =- 500 - 1,000 - 2,000 - 200x6=-4,700, 
y 7 =- 500 - 1,000 - 2,000 - 200 X 7=-4,900, 
y 8 =- 500 - 1,000 - 2,000 - 200x8=-5,100, 
y 9 =-500 - 1,000 - 2,000 - 200x9=-5,300. 

The values, being negative, should be plotted downward. To a 
scale of 5,000 pounds to the inch they give the shear lines aft. Vc. 
c'd (Fig. 16, ft). 

EXAMPLES FOR PRACTICE. 

1. Construct a shear diagram for the beam represented in 
Fig. 10, neglecting the weight of the beam (see example 1, Art. 37). 

2. Construct the shear diagram for the beam represented in 
Fig. 11, neglecting the weight of the beam (see example 3, 
Art, 37). 


















































































STRENGTH OF MATERIALS 


O 1 

O 1 

3. Construct the shear diagram for the beam of Fig. 12 
when it sustains, in addition to the loads represented, its own 
weight, 800 pounds, and a uniform load of 500 pounds per foot 
(see example 4, Art. 37). 

4. Figs, a, cases 1 and 2, Table B, represent two cantilever 
beams, the first bearing a concentrated load P at the free end, 
and the second a uniform load W. Figs, b are the corresponding 
shear diagrams. Take P and W ecpial to 1,000 pounds, and satisfy 
yourself that the diagrams are correct. 

and satisfy yourself that the diagrams are correct. 

5. Figs, a, cases 3 and 4, same table, represent simple 
beams supported at their ends, the first bearing a concentrated 



c 


Sca-le i" = 5 ooo lbs. d 

Fig. 16. 

load P at the middle, and the second a uniform load W. Figs. 
b are the corresponding shear diagrams. Take P and W equal 
to 1,000 pounds, and satisfy yourself that they are correct. 

39. Maximum Shear. It is sometimes desirable to know 
the greatest or maximum value of the shear in a given case. This 
value can always be found with certainty by constructing the shear 
diagram, from which the maximum value of the shear is evident at 
a glance. In any case it can most readily be computed if one 
knows the section for which the shear is a maximum. The stu¬ 
dent should examine all the shear diagrams in the preceding 
articles and those that he has drawn, and see that 

1. In cantilevers fixed in a wall , the maximum shear 

occurs at the wall . 














































































32 


STRENGTH OF MATERIALS 


2. In simple beams , the maximum shear occurs at a sec¬ 
tion next to one of the supports. 

By the use of these propositions one can determine the value 
of the maximum shear without constructing; the whole shear 
diagram. Thus, it is easily seen (referring to the diagrams, page 
53) that for a 

Cantilever, end load P, maximum shear=P 
“ , uniform load W, “ “ =W 

Simple beam, middle load P, “ “ =JP 

“ “ , uniform “ W, “ “ = ^W 

40. Bending Homent. By bending moment at (or for) a 
section of a loaded beam, is meant the algebraic sum of the mo¬ 
ments of all the loads (including weight of beam) and reactions 
to the left or right of the section with respect to any point in the 
section. 

41. Rule of Signs. We follow the rule of signs previously 
stated (Art. 29) that the moment of a force which tends to pro¬ 
duce clockwise rotation is plus, and that of a force which tends to 
produce counter-clockwise rotation is minus; but in order to get 
the same sign for the bending moment whether computed from 
the right or left, we change the sign of the sum of the moments 
when computed from the loads and reactions on the right. Thus 
for section a, Fig. 8, the algebraic sums of the moments of the 
forces are: 

when computed from tne left, 

—1,000X5 + 3,000X1=-2,000 foot-pounds; 
and when computed from the right, 

1,000 X 19-3,000 X15 ■+ 2,000 X13 + 2,000 X1=■+ 2,000 foot- 

The bending moment at section a is -2,000 foot-pounds. 

Again, for section b, the algebraic sums of the moments of the 
forces are: 

when computed from the left, 

-1,000 X 22 + 3,000 X 18-2,000 X 16-2,000 X 4+3,000 X 2= 
-2,000 foot-pounds; 
and when computed from the right, 

1,000x2=+ 2,000 foot-pounds. 

The bending moment at the section is -2.000 foot-pounds. 



STRENGTH OF MATERIALS 


QQ 

GO 


It is usually convenient to compute the bending moment for 
a section from the forces to the right or left according as there 
are fewer forces (loads and reactions) on the right or left side 
of the section. 

42. Units. It is customary to express bending moments in 
inch-pounds, but often the foot-pound unit is more convenient. 
To reduce foot-pounds to inch-pounds , multiply by twelve. 

43. Notation. We shall use M to denote bending moment at 
any section, and the bending moment at a particular section will 
be denoted by that letter subscripted; thus M 2 , etc., denote 
values of the bending moment for sections one, two, etc., feet 
from the left end of the beam. 

Examples. 1. Compute the bending moments for sections 
one foot apart in the beam represented in Fig. 9, neglecting the 
weight of the beam. (The right and left reactions are 3,700 and 
2,300 pounds respectively. See example 1, Art. 33.) 

Since there are no forces acting on the beam to the left of the 
left support, M o =0. To the left of the section one foot from the 
left end there is but one force, the left reaction, and its arm is one 
foot; hence M. 1 = + 2,300X 1=2,300 foot-pounds. To the left of 
a section two feet from the left end there are two forces, 2,300 and 
1,000 pounds, and their arms are 2 feet and 1 foot respectively; 
hence M 2 = + 2,300 X 2-1,000 X 1=3,600 foot-pounds. At the 
left of all sections between B and C there are only two forces, 
2,300 and 1,000 pounds; hence 

M 3 = + 2,300 X 3-1,000 X 2= + 4,900 foot-pounds, 

M 4 = + 2,300 X 4-1,000 X 3= + 6,200 
M 5 = + 2,300 X 5-1,000 X 4= + 7,500 
M 6 = + 2,300 X 6-1,000 x 5= + 8,800 

To the right of a section seven feet from the left end there 
are two forces, the 3,000-pound load and the right reaction 
(3,700 pounds), and their arms with respect to an origin in that 
section are respectively one foot and three feet; hence 

M 7 =-(-3,700 X 3 + 3,000 X 1)= + 8,100 foot-pounds. 

To the right of any section between E and D there is only one 
force, the right reaction; hence 


34 


STRENGTH OF MATERIALS 


M 8 =-(-3,700 X 2)=7,400 foot-pounds, 

M 9 =-(-3,700 X 1)=3,700 

Clearly M 10 =0. 

2. A simple beam 10 feet long and supported at its ends 
weighs 400 pounds, and bears a uniformly distributed load of 1,000 

Compute the bending moments for sections two feet 

Each reaction equals one-half the whole load, that is, -J of 
(1,600 + 400) =1,000 pounds, and the load per foot including 
weight of the beam is 200 pounds. The forces acting on the 
beam to the left of the first section, two feet from the left end, are 
the left reaction (1,000 pounds) and the load (including weight) 
on the part of the beam to the left of the section (400 pounds). 
The arm of the reaction is 2 feet and that of the 400-pound force 
is 1 foot (the distance from the middle of the 400-pound load to 
the section). Hence 

Mo—+ 1,000 X 2-400 X 1 = +1,600 foot-pounds. 

The forces to the left of the next section, 4 feet from the left 
end, are the left reaction and all the load (including weight of 
beam) to the left (800 pounds). The arm of the reaction is 4 feet, 
and that of the 800-pound force is 2 feet; hence 

M 4 = +1,000 X 4-800 X 2= + 2,400 foot-pounds. 

Without further explanation the student should see that 

M 6 = +1,000 X 6-1,200 X 3= + 2,400 foot-pounds, 

M 8 = +1,000 X 8-1,600 X4= +1,600 

Evidently M 0 =M 10 =0. 

3. Compute the values of the bending moment in example 
1, taking into account the weight of the beam, 400 pounds. (The 
right and left reactions are respectively 3,900 and 2,500 pounds; 
see example 3, Art. 33.) 

We proceed as in example 1, except that the moment 
of the weight of the beam to the left of each section (or to 
the right when computing from forces to the right) must be 
included in the respective moment equations. Thus, computing 
from the left, 



35 


STRENGTH OE MATERIALS 


M 0 =0 

Mi — -j- 2,500 X 1-40 X £= + 2,480 foot-pounds, 

M 2 = + 2,500 X 2-1,000 X 1-80 X1=■+ 3,920, 

M 3 = + 2,500 X 3-1,000 X 2-120 X 1£= + 5,320, 

M 4 = + 2,500 X 4-1,000 X 3-160 X 2=+ 6,680, 

M 5 = + 2,500 X 5-1,000 X 4-200 X 2±= + 8,000, 

M 6 = + 2,500 X 6-1,000 X 5-240 X 3=+9,280. 

Computing from the right, 

M 7 =-(-3,900 X 3 + 3,000 X1 +120 X14) = + 8,520, 
M 8 =-(-3,900x2 + 80x1) = + ?,720, " 

M 9 =-(-3,900x1 + 40x4)=+ 3,880, 

M 10 = 0. 

EXAMPLES FOR PRACTICE. 


1. Compute the values of the bending 
one foot apart, beginning one foot from 
beam represented in Fig. 10, neglecting the 
(The right and left reactions are 3,300 and 
tively; see example 2, Art. 33.) 

f M x = - 2,100 M r = + 3,400 M n = 
Ans. M 2 = - 4,200 M 7 = + 5,300 M 
(in foot- - M 3 = - 2,300 M 8 = + 7,200 M 
pounds) M 4 = - 400 M 9 = + 5,500 M 
M 5 = +1,500 M 10 = +3,800 M„: 


moment for sections 
the left end of the 
weight of the beam. 
4,000 pounds respec- 


- 12 ' 


- 13 ' 


- 14 ' 


+ 2,100 M 16 = 
+ 400 M„ 

-1,300 M 18 = 

- 3,000 M 19 = 

- 4,700 M 20 = 


-6,400 

-4,800 

-3,200 

1,600 


0 


2. Solve the preceding example, taking into account the 
weight of the beam, 42 pounds per foot. (The right and left 
reactions are 3,780 and 4,360 pounds respectively; see example 4, 
Art. 33.) 


Ans. 
(in foot¬ 
pounds) 


■< 


M,= 
M = 
M = 
M = 
M = 


- 2,121 M, 

- 4,284 M, 

- 2,129 M 8 

16 M, 
+ 2,055 M 


11 ' 


10 ‘ 


+4,084 M 
:+6,071 M„ 
:+8,016 M„ 
+ 6,319 M u 
+ 4,580 M, 


+ 2,799 M„ 
+ 976 M„ 

- 889 M 

- 2,796 M 

- 4,745 M, 


18 


19 


- 6,736 

- 4,989 

- 3,284 

- 1,621 

0 


3. Compute the bending moments for sections one foot 
apart, of the beam represented in Fig. 11, neglecting the weight. 
(The right and left reactions are 1,444 and 1,556 pounds respect- 
ively; see example 1, Art. 33.) 






36 


STRENGTH OF MATERIALS 


r 


Ans. 

(in foot- - 
pounds) 


M,= 

M 2 : 
M,= 

M,: 


+ 1,556 M 5 = 
=+3,112 M„= 
=+4,068 M,= 
+ 5,024 M = 


+ 5,980 M„ 
+6,936 M 10 
+ 6,992 M„ 
+6,548 M ls 


13 ' 


+ 6,104 M 
+ 5,660 M u 
+ 5,216 M i5 
+ 4,772 M 16 


+ 4,328 
+ 2,884 
+ 1,440 
0 


4 Compute the bending moments at sections one foot apart 
in the beam of Fig. 12, taking into account the weight of the beam, 
800 pounds, and a uniform load of 500 pounds per foot. (The 
right and left reactions are 4,870 and 11,930 pounds respectively; 
see Exs. 3 and 4, Art. 33.) 


Ans. 

(in foot- -= 
pounds) 


270 M 6 = -19,720 M u =+ 3,980 M 1$ =12,180 
M a =- 3,080 M 7 = -13,300 M ia = + 6,700 M 17 =12,200 
M 8 =- 6,430 M 8 = - 7,420 M 1S =+ 8,880 M 18 = 8,680 
M 4 = -10,320 M 9 =- 3,080 M 14 =: +10,520 M 19 == 4,620 
M 5 = -14,750 M 10 = + 720 M 15 = +11,620 M 20 = 0 


44. Moment Diagrams. The way in which the bending 
moment varies from section to section in a loaded beam can be 
well represented by means of a diagram called a moment diagram. 
To construct such a diagram for any loaded beam, 



Fig. 17. 


1. Lay off a base-line just as for a shear diagram (see 
Art. 38). 

2. Draw a line such that the distance from any point of it 
to the base-line equals (by some scale) the value of the bending 
moment at the corresponding section of the beam, and so that the 
line is above the base where the bending moment is positive and 
below it where it is negative. (This line is called a “moment 
line/*) 






























































































































STRENGTH OF MATERIALS 


37 


Examples. 1. It is required to construct a moment dia¬ 
gram for the beam of Fig. 17, a (a copy of Fig. 9), loaded as 
there shown. 

Layoff A'E' (Fig. 17, b) as a base. In example 1, Art. 43, 
we computed the values of the bending moment for sections one 
foot apart, so we erect ordinates at points of A'E' one foot apart, 
to represent the bending moments. 

We shall use a scale of 10,000 foot-pounds to the inch; then 
the ordinates (see example 1, Art. 43, for values of M) will be: 
One foot from left end, 2,300-^-10,000 = 0.23 inch, 

Two feet “ “ “ 3,600-^10,000 = 0.36 “ 

Three “ “ “ “ 4,900-7-10,000 = 0.49 “ 

Four “ “ u “ 6,200-f-10,000 = 0.62 “ 

etc., etc. 


looolba. 2ooolbs. 3000 lb 3 . 



2s. 


b 


Laying these ordinates off, and joining their ends in succession, 
we get the line A'bcdE\ which is the bending moment line. 
Fig. 17, b, is the moment diagram. 

2. It is required to construct the moment diagram for the 
beam, Fig. 18, a (a copy of Fig. 9), taking into account the weight 
of the beam, 400 pounds. 

The values of the bending moment for sections one foot apart 
were computed in example 3, Art. 43. So we have only to lay off 
ordinates equal to those values, one foot apart, on the base A'E : 
(Fig. 18, l). 

To a scale of 10,000 foot-pounds to the inch the ordinates 
(see example 3, Art. 43, for values of M) are: 





















































































































38 


STRENGTH OF MATERIALS 


At left end, 0 

One foot from left end, 2,480-5-10,000=0.248 inch 
Two feet 44 44 44 3,920-5-10,000=0.392 44 

Three 44 44 44 “ 5,320-5-10,000=0.532 “ 

Four 44 44 44 44 6,680-5-10,000=0.668 44 

Laying these ordinates off at the proper points, we get A'bcdE 
as the moment line. 

3. It is required to construct the moment diagram for the 
cantilever beam represented in Fig. 19, a , neglecting the weight 
of the beam. The bending moment at B equals 
-500X 2=-l,000 foot-pounds; 

at C, 

-500 X 5-1,000 X 3=-5,500; 

and at D, 

-500 X 9-1,000 X 7-2,000 X 4=-19,500. 


500 lbs. looolbs. aooo lbs. 



Using a scale of 20,000 foot-pounds to one inch, the ordinates 
in the bending moment diagram are: 

At B, 1,000^-20,000=0.05 inch, 

44 C, 5,500-v-20,000=0.275 44 
44 D, 19,500-5-20,000=0.975 44 

Hence we lay these ordinates off, and downward because the bend¬ 
ing moments are negative, thus fixing the points Z>, c and d. The 
bending moment at A is zero; hence the moment line connects A 
Z>, c and d. Further, the portions A4, 1)0 and cd are straight, as 
can be shown by computing values of the bending moment for 
sections in AB, BO and CD, and laying off the corresponding 
ordinates in the moment diagram. 

























































STRENGTH OF MATERIALS 


39 


4. Suppose that the cantilever of the preceding illustration 
sustains also a uniform load of 100 pounds per foot (see Fig. 20, a). 
Construct a moment diagram. 

ci 

First, we compute the values of the bending moment at sev¬ 
eral sections; thus, 

M t =-500 X1-100 X J=-550 foot-pounds, 

M 2 =-500 X 2-200 X1=-1,200, 

M 3 =-500 X 3-1,000 X1-300 X l±=-2,950, 

M 4 =-500 X 4-1,000 X 2-400 X 2=-4,800, 

M 5 =-500 X 5-1,000 X 3-500 X 2£=-6,750, 

M u =-500 X 6-1,000 X 4-2,000 X1-600 X 3=-10,800, 

M ? =-500 X 7-1,000 X 5-2,000 X 2-700 X 3J=-14,950, 
M 8 =-500 X 8-1,000 X 6-2,000 X 3-800 X 4=-19,200, 

M 9 =-500 X 9-1,000 X 7-2,000 X 4-900 X 4£=-23,550. 

500 lbs. looolbs. 2000 lbs. 

dc 


"to 


Fig. 20. 

These values all being negative, the ordinates are all laid off 
downwards. To a scale of 20,000 foot-pounds to one inch, they 
fix the moment line A 'bed. 

EXAHPLES FOR PRACTICE. 

1. Construct a moment diagram for the beam represented in 
Fig. 10, neglecting the weight of the beam. (See example 1, 
Art. 43). 

2. Construct a moment diagram for the beam represented 
in Fig. 11, neglecting the weight of the beam. (See example 3, 
Art. 43). 

3. Construct the moment diagram for the beam of Fig. 12 























































40 


STRENGTH OF MATERIALS 


when it sustains, in addition to the loads represented and its own 
weight (800 pounds), a uniform load of 500 pounds per foot. 
(See example 4, Art. 43.) 

4. Figs, a , cases 1 and 2, page 53, represent two cantilever 
beams, the first bearing a load P at the free end, and the second 
a uniform load W. Figs, c are the corresponding moment 
diagrams. Take P and W equal to 1,000 pounds, and l equal to 
10 feet, and satisfy yourself that the diagrams are correct. 

5. Figs. a, cases 3 and 4, page 53, represent simple beams 
on end supports, the first bearing a middle load P, and the other a 
uniform load W. Figs, c are the corresponding moment dia¬ 
grams. Take P and W equal to 1,000 pounds, and l equal to 
10 feet, and satisfy yourself that the diagrams are correct. 

45. Maximum Bending Moment. It is sometimes desirable 
to know the greatest or maximum value of the bending moment 
in a given case. This value can always be found with certainty 
by constructing the moment diagram, from which the maximum 
value of the bending moment is evident at a glance. But in any 
case, it can be most readily computed if one knows the section for 
which the bending moment is greatest. If the student will com¬ 
pare the corresponding shear and moment diagrams which have 
been constructed in foregoing articles (Figs. 13 and 17, 14 and 
18, 15 and 19, 16 and 20), and those which he has drawn, he will 
see that —The maximum bending moment in a beam occurs 
where the shear changes sign. 

By the help of the foregoing principle we can readily com¬ 
pute the maximum moment in a given case. We have only to 
construct the shear line, and observe from it where the shear 
changes sign; then compute the bending moment for that section. 
If a simple beam has one or more overhanging ends, then the shear 
changes sign more than once—twice if there is one overhanging 
end, and three times if two. In such cases we compute the 
bending moment for each section where the shear changes sign; 
the largest of the values of these bending moments is the maxi¬ 
mum for the beam. 

The section of maximum bending moment in a cantilever 
fixed at one end (as when built intD a wall) is always at the wall. 


STRENGTH OF MATERIALS 


41 


Thus, without reference to the moment diagrams, it is readily seen 
that, 

for a cantilever whose length is L 

with an end load P, the maximum moment is P l, 

44 a uniform 44 AY, 44 44 44 44 Wl. 

Also by the principle, it is seen that, 

for a beam whose length is l , on end supports, 

with a middle load P, the maximum moment is ^ P/, 

44 uniform 44 AY, 44 44 44 44 1 A WL 

46. Table of Maximum Shears, Moments, etc. Table B 
on page 53 shows the shear and moment diagrams for eight 
simple cases of beams. The first two cases are built-in cantilevers; 
the next four, simple beams on end supports; and the last two, 
restrained beams built in w r alls at each end. In each case l 
denotes the length. 

CENTER OF GRAVITY AND HOMENT OF INERTIA. 

It will be shown later that the strength of a beam depends 
partly on the form of its cross-section. The following discussion 
relates principally to cross-sections of beams, and the results 
reached (like shear and bending moment) will be made use of 
later in the subject of strength of beams. 

47. Center of Gravity of an Area. The student probably 
knows what is meant by, and how to find, the center of gravity of 
any flat disk, as a piece of tin. Probably his way is to balance 
the piece of tin on a pencil point, the point of the tin at wdiich it so 
balances being the center of gravity. (Really it is midway between 
the surfaces of the tin and over the balancing point.) The center 
of gravity of the piece of tin, is also that point of it through which 
the resultant force of gravity on the tin (that is, the weight of the 
piece) acts. 

By 44 center of gravity” of a plane area of any shape we mean 
that point of it which corresponds to the center of gravity of a 
piece of tin when the latter is cut out in the shape of the area. 
The center of gravity of a quite irregular area can be found most 
readily by balancing a piece of tin or stiff paper cut in the shape 
of the area. But wdien an area is simple in shape, or consists of 
parts which are simple, the center of gravity of the whole can be 


42 


STRENGTH OF MATERIALS 


found readily by computation, and such a method will now be 

described. 


48. Principle of floments Applied to Areas. Let Fig. 21 


represent a piece of tin which has been divided off into any num¬ 
ber of parts in any way, the weight of the whole being W, and 
that of the parts W 1? W 2 , W 3 , etc. Let C n C 2 , C 3 , etc., be the 
centers of gravity of the parts, C that of the whole, and <?„ c 2 , c 3 , 
etc., and c the distances from those centers of gravity respectively 


to some line (L L) in the plane 
of the sheet of tin. When the 
tin is lying in a horizontal posi¬ 
tion, the moment of the weight 
of the entire piece about L L is 
W c, and the moments of the 
parts are W&, W 2 <? 2 , etc. Since 
the weight of the whole is the 
resultant of the weights of the 
parts, the moment of the weight 
of the whole equals the sum of t 
parts; that is, 



moments of the weights of the 


"W c —"W^Cj-j-W .... 


Now let Aj, A 2 , etc. denote the areas of the parts of the pieces 
of tin, and A the area of the whole; then since the weights are 
proportional to the areas, we can replace the W’s in the preceding 
equation by corresponding A’s, thus: 

Ac= AjCj + A 2 c 2 +etc. ( 4 ) 

If we call the product of an area and the distance of its 
center of gravity from some line in its plane, the “moment” of the 
area with respect to that line, then the preceding equation may be 
stated in words thus: 

The moment of an area with respect to any line equals the 
algebraic sum of the moments of the parts of the area. 

If all the centers of gravity are on one side of the line with 
respect to which moments are taken, then all the moments should be 
given the plus sign; but if some centers of gravity are on one side 
and some on the other side of the line, then the moments of the 
areas whose centers of gravity are on one side should be given the 








STRENGTH OF MATERIALS 


43 


same sign, and the moments of the others the opposite sign. The 
foregoing is the principle of moments for areas, and it is the basis 
of all rules for finding the center of gravity of an area. 

To find the center of gravity of an area which can be divided 
up into simple parts, we write the principle in forms of equations 
for two different lines as “axes of moments,” and then solve the 
equations for the unknown distances of the center of gravity of the 
whole from the two lines. We explain further by means of specific 
examples. 

Examples. 1. It is required to find the center of gravity 
of Fig. 22, a, the width being uniformly one inch. 

The area can be divided into two rectangles. Let 0, and 

C3 * 




oh -a-—H 





-3: 

c. 




• 1 

•C 

C S 


_ 

•c 2 

•j 

\ f 

-12 ,L -* 

r ~ 




To 

Fig. 22. 


Co be the centers of gravity of two such parts, and C the center of 
gravity of the whole. Also let a and 6 denote the distances of C 
from the two lines OL' and OL" respectively. 

The areas of the parts are G and 3 square inches, and their 
arms with respect to OL' are 4 inches and inch respectively, and 
with respect to OL" -J inch and 14 inches. Hence the equations of 
moments with respect to OL' and OL" (the whole area being 9 


square inches) 

are: 




9X« 

=6X4+3X|= 

25.5, 


9X6 

= 6X^+3xlg = 

= 7.5. 

Hence, 

a —- 

25.5-r-9 — 2.83 inches, 


6 — 

7.5-h9 = 0.83 

u 

• 


2. It is required to locate the center of gravity of Fig. 22, 6, 
the width being uniformly one inch. 
























44 


STRENGTH OF MATERIALS 


The figure can be divided up into three rectangles. Let C 1? C 2 
and C.j be the centers of gravity of such parts, C the center of 
gravity of the whole; and let a denote the (unknown) distance of 
C from the base. The areas of the parts are 4, 30 and 4 square 
inches, and their “ arms ” with respect to 
the base are 2 ,^ and 2 inches respectively; 
hence the equation of moments with re¬ 
spect to the base (the entire area being 18 
square inches) is: 

18 X & 4x2 +10 X i + 4 X 2 = 21. 

Hence, a — 21 -r- 18 = 1.17 inches. 


M" 


-r 


From the symmetry of the area it is plain 
that the center of gravity is midway be¬ 
tween the sides. 

EXAMPLE FOR PRACTICE. 

1. Locate the center of gravity of 



Fig. 23. Fig. 23. 

Ans. 2.3 inches above the base. 

49 . Center of Gravity of Built=up Sections. In Fig. 24 

there are represented cross-sections of various kinds of rolled steel, 
called “shape steel,” which is used extensively in steel construction. 
Manufacturers of this material publish “handbooks” giving full 
information in regard thereto, among other things, the position of 
the center of gravity of each cross section. With such a handbook 



Channel 



T-ba^r 


Fig. 24. 


available, it is therefore not necessary actually to compute the posi- 
tion of the center of gravity of any section, as we did in the pre¬ 
ceding article; but sometimes several shapes are riveted together to 





































STRENGTH OF MATERIALS 


45 


make a “built-up” section (see Fig. 25), and tlien it may be neces¬ 
sary to compute the position of the center of gravity of the section. 

Example. It is desired to locate the center of gravity of the 
section of a built-up beam represented in Fig. 25. The beam con- 



Fig. 25. 

sists of two channels and a plate, the area of the cross-section of a 
channel being 6.03 square inches. 

Evidently the center of gravity of each channel section is 6 
inches, and that of the plate section is 12^ inches, from the bottom. 

Let c denote the dis- 


4 


1 


e*, 


b 


h. 




66 " 


tance of the center of 
gravity of the whole 
section from the bot¬ 
tom; then since the 
area of the plate section 
is 7 square inches, and 
that of the whole sec¬ 
tion is 19.06, 


Hence, 


Fig. 26. 

c= 158.11 


19.06X^ = 6.03 X 6 + 
6.03 X6 + 7X12| = 
158.11. 

-^-19.06=8.30 inches, (about). 


EXAMPLES FOR PRACTICE. 

1. Locate the center of gravity of the built-up section of 































4G 


STRENGTH OF MATERIALS 


Fig. 26, a, the area of each “angle” being 5.06 square inches, and 
the center of gravity of each being as shown in Fig. 26, b. 

Ans. Distance from top, 3.08 inches. 
2. Omit the left-hand angle in Fig. 26, a, and locate the 


center of gravity of the remainder. 

. ( Distance from top, 3.65 inches, 

nS * } “ “ left side, 1.19 inches. 

50. rioment of Inertia. If a plane area be divided into an 
infinite number of infinitesimal parts, then the sum of the prod¬ 
ucts obtained by multiplying the area of each part by the square 
of its distance from some line is called the moment of inertia of the 
area with respect to the line. The line to wdiich the distances are 
measured is called the inertia-axis; it may be taken anywhere in 
the plane of the area. In the subject of beams (where we have 
sometimes to compute the moment of inertia of the cross-section 
of a beam), the inertia-axis is taken through the center of gravity 
of the section and horizontal. 

An approximate value of the moment of inertia of an area 
can be obtained by dividing the area into small parts (not infini¬ 
tesimal), and adding the products obtained by multiplying the 
area of each part by the square of the distance from its center to 
the inertia-axis. 

Example. If the rectangle of Fig. 27, a , is divided into 8 
parts as shown, the area of each is one square inch, and the dis¬ 
tances from the axis to the centers of gravity of the parts are ^ 
and 1^ inches. For the four parts lying nearest the axis the 
product (area times distance squared) is: 


1X( -|) 2 =J; an( ^ f° r the other parts it is 

Hence the approximate value of the moment of inertia of the area 
with respect to the axis, is 

4 (i)+4(f)=10. 

If the area is divided into 32 parts, as shown in Fig. 27, b , 
the area of each part is J square inch. For the eight of the little 
squares farthest away from the axis, the distance from their centers 
of gravity to the axis is 1| inches; for the next eight it is 1^; 
for the next eight and for the remainder ^ inch. Hence an 


STRENGTH OE MATERIALS 


47 


approximate value of the moment of inertia of the rectangle with 
respect to the axis is: 


8XiX(l|) 2 +8xJX(li) 2 +8 X JX(|) 2 +8xiX(i) 2 =10J. 

If we divide the rectangle into still smaller parts and form 
the products 

(small area) X (distance) 2 , 

and add the products just as we have done, we shall get a larger 
answer than 10J. The smaller the parts into which the rec¬ 
tangle is divided, the larger will be the answer, but it will never 
be larger than 10J. This 10§ is the sum corresponding to a 

division of the rectangle into an 
infinitely large number of parts 
(infinitely small) and it is the 
exact value of the moment of 
inertia of the rectangle with re¬ 
spect to the axis selected. 

There are short methods of 
computing the exact values of the 
moments of inertia of simple fig¬ 
ures (rectangles, circles, etc.,), 
but they cannot be given here since they involve the use of difficult 
mathematics. The foregoing method to obtain approximate val¬ 
ues of moments of inertia is used especially when the area is quite 
irregular in shape, but it is given here to explain to the student 
the meaning of the moment of inertia of an area. He should 
understand now that the moment of inertia of an area is sim¬ 
ply a name for such sums as we have just computed. The name 
is not a fitting one, since the sum has nothing whatever to do with 
inertia. It was first used in this connection because the sum is 
very similar to certain other sums which had previously been 
called moments of inertia. 

51. Unit of Moment of Inertia. The product (areaX dis¬ 
tance 2 ) is really the product of four lengths, two in each factor; 
and since a moment of inertia is the sum of such products, a 
moment of inertia is also the product of four lengths. Now the 
product of two lengths is an area, the product of three is a vol¬ 
ume, and the product of four is moment of inertia— unthinkable in 



to 

Fig. 27. 



























48 


STRENGTH OF MATERIALS 


the way in which we can think of an area or volume, and there¬ 
fore the source of much difficulty to the student. The units of 
these quantities (area, volume, and moment of inertia) are respec- 

the square inch, square foot, etc., 

“ cubic “ , cubic “ “ , 

“ biquadratic inch, biquadratic foot, etc.; 

but the biquadratic inch is almost exclusively used in this connec¬ 
tion; that is, the inch is used to compute 
values of moments of inertia, as in the pre¬ 
ceding illustration. It is often written 

thus: Inches 4 . 

52. Moment of Inertia of a Rectangle. 

Let b denote the base of a rectangle, and a 
its altitude; then by higher mathematics it 
can be shown that the moment of inertia 
of the rectangle with respect to a line through its center of gravity 
and parallel to its base, is y * 1 * ^ bo .A 

Example . Compute the value of the moment of inertia of 
a rectangle 4x12 inches with respect to a line through its center 
of gravity and parallel to the long side. 

Here b= 12, and a = 4 inches ; hence the moment of inertia 
desired equals 

-J^(12X4 3 )=64 inches 4 . 

EXAHPLE FOR PRACTICE. 

1. Compute the moment of inertia of a rectangle 4x12 

inches with respect to a line through its center of gravity and 
parallel to the short side. Ans. 576 inches 4 . 

53. Reduction Formula. In the previously mentioned 
“handbooks” there can be found tables of moments of inertia of 
all the cross-sections of the kinds and sizes of rolled shapes made. 
The inertia-axes in those tables are always taken through the cen¬ 
ter of gravity of the section, and usually parallel to some edge of 
the section. Sometimes it is necessary to compute the moment of 
inertia of a “rolled section” with respect to some other axis, and 
if the two axes (that is, the one given in the tables, and the other) 
are parallel, then the desired moment of inertia can be easily com¬ 
puted from the one given in the tables by the following rule: 


T 

Oi 


8 "- 


T 


axis 


Fig, 28. 












STRENGTH OF MATERIALS 


49 


The moment of inertia of an area with respect to any axis 
equals the moment of inertia with resqiect to a parallel axis 
through the center of gravity, plus the product of the area and 
the square of the distance between the axes. 

Or. if I denotes the moment of inertia with respect to any axis; 
I Q the moment of inertia with respect to a parallel axis through 
the center of gravity; A the area; and d the d : stance between the 
axes, then 

I—+ Ad 2 . . . . (q;) 

Example. It is required to compute the moment of inertia 
of a rectangle 2 x 8 inches with respect to a line parallel to the 
long side and 4 inches from the center of gravity. 

Let I denote the moment of inertia sought, and I Q the moment 
of inertia of the rectangle with respect 
to a line parallel to the long side and 
through the center of gravity (see Fig. 

28). Then 

I Q — (see Art. 52); and, 

since 3=8 inches and a=2 inches, 

I 0 = t 1 2-(8 X 2 3 ) = 5J biquadratic inches. 

The distance between the two inertia- 
axes is 4 inches, and the area of the 
rectangle is 16 square inches, hence 
equation 5 becomes 

I=5J-fl6 X4 2 =261J biquadratic inches. 

EXAMPLE FOR PRACTICE* 

1. The moment of inertia of an “angle” 2^X2x|- inches 
(lengths of sides and width respectively) with respect to a line 
through the center of gravity and parallel to the long side, is 0.64 
inches 4 . The area of the section is 2 square inches, and the dis¬ 
tance from the center of gravity to the long side is 0.63 inches. 
(These values are taken from a “handbook”.) It is required to 
compute the moment of inertia of the section with respect to a 
line parallel to the long side and 4 inches from the center of 
gravity. Ans. 32.64 inches 4 . 

54. Moment of Inertia of Built-up Sections. As before 
stated, beams are sometimes “built up” of rolled shapes (angles, 


ocis 


c—4"_*| 

—*- 


CO 


H —>1 




Fig. 29. 



















50 


STRENGTH OF MATERIALS 


channels, etc.). The moment of inertia of such a section with 
respect to a definite axis is computed by adding the moments of 
inertia of the parts, all with respect to that same axis. This is the 
method for computing the moment of any area which can be 
divided into simple parts. 

The moment of inertia of an area which may be regarded as 
consisting of a larger area minus other areas, is computed by sub¬ 
tracting from the moment of inertia of the large area those of the 
“minus areas.” 

Examples. 1. Compute the moment of inertia of the built- 
up section represented in Fig. 30 (in part same as Fig. 25) with 
respect to a horizontal axis 
passing through the center . f- 



—14- 


- 4 ™ 


Fig. 30. 


iL-e? 


o 

<7 

OJ 

Tt 


■til 


o 

1 


of gravity, it being given 
that the moment of inertia 
of each channel section 
with respect to a horizontal 
axis through its center of 
gravity is 128.1 inches 4 , 
and its area 6.03 square 
inches. 

The center of gravity of 
the whole section was found 

in the example of Art. 49 to be 8.30 inches from the bottom of 
the section; hence the distances from the inertia-axis to the 
centers of gravity of the channel section and the plate are 2.30 
and 3.95 inches respectively (see Fig. 30). 

The moment of inertia of one channel section with respect to 
the axis A A (see equation 5, Art. 53) is: 


128.1 + 6.03 X2.30 2 =160.00 inches 4 . 

The moment of inertia of the plate section (rectangle) with re¬ 
spect to the line a'a' (see Art. 52) is: 

tV ba 3 =q!^[14X (-|) 3 ]—0.15 inches 4 ; 
and with respect to the axis A A (the area being 7 square inches) 
it is: 

0.15+7X3.95 2 =109.37 inches 4 . 

Therefore the moment of inertia of the whole section with re- 
spect to AA is: 

2x160.00 + 109.37=429.37 inches 4 . 
























STRENGTH OF MATERIALS 


51 


2. It is required to compute tlie moment of inertia of the 
“ hollow rectangle ” of Fig. 29 with respect to a line through the 
center of gravity and parallel to the short side. 

The amount of inertia of the large rectangle with respect to 
the named axis (see Art. 52) is: 

T V (5 X10 3 ) = 416§; 


B 



same axis is: 

tV (4X8°) = 170§; 

hence the moment of inertia of the hollow section with respect 
to the axis is: 

416§ - 170§ = 246 inches 4 . 

EXAMPLES FOR PRACTICE. 

1. Compute the moment of inertia of the section repre¬ 

sented in Fig. 31, about the axis AA, it being 3.08 inches 
from the top. Given also that the area of one angle section is 
5.06 square inches, its center of gravity C (Fig. 31, b) 1.66 inches 
from the top, and its moment of inertia with respect to the axis aa 
17.68 inches 4 . Ans. 145.8 inches 4 - 

2, Compute the moment of inertia of the section of Fig. 31, a, 


























52 


STRENGTH OF MATERIALS 


with respect to the axis BB. Given that distance of the center 
of gravity of one angle from one side is 1.66 inches (see Fig. 31, £), 
and its moment of inertia with respect to bb 17.68 inches. 

Ans. 77.618 inches 4 . 

55. Table of Centers of Gravity and rioments of Inertia. 

Column 2 in Table A below gives the formula for moment of 
inertia with respect to the horizontal line through the center of 
gravity. The numbers in the third column are explained in Art. 
62; and those in the fourth, in Art. 80. 

TABLE A. 

Moments of Inertia, Section Moduli, and Radii of Gyration. 

In each case the axis is horizontal and passes through the center of gravity. 


Section. 




Moment of 
Inertia. 


a* 

12 


a 4 - aj 4 
12 

ba 3 

12 


ba 8 -b ] a, 3 

12 


0.049d 4 


0.049 (d 4 - dj 4 ) 


Section 

Modulus. 


a 5 

6 

a 4 -a/ 
6a 

ba 2 

6 


ba^-b ^ 3 

6a 


0.098d 3 


0.098- 


d 4 -d 4 


Radius of 
Gyration. 


a 


V 12 


4 


a 2 -f- a^ 
~12 


a 


j/ 12 


>1 


ba 3 -b 1 a 1 3 
12 (ba - b^) 

d 


]/d 2 + d, 2 


STRENGTH OF BEAMS. 

56. Kinds of Loads Considered. The loads that are applied 

to a horizontal beam are usually vertical, but sometimes forces are 
applied otherwise than at right angles to beams. Forces acting on 
beams at right angles are called transverse forces; those applied 






























































STRENGTH OF MATERIALS 


53 


TABLE B. 

Shear Diagrams (b) and Moment Diagrams (c) for Eight Different Cases (a). 
Also Values of Maximum Shear (V), Bending floment (M), and Deflection (d). 


1 f 

e* *- 


X 

W-uml 

"orm ioto.d 

6.. i-- 


--i- 


- --1 — 


to TMI 1 III 

10 .mill.. 



lllllli 


V=P, M=P1, d= 

P1 3 +3EI. 

V=W, M=% Wl, 

d=WI 3 -s-8EI. 

3 f 

4 

W-uniform loa-d 

A 

.T. 

. . .1.. 

A 

to- A - 

, innniiiiiTiTnmTtm^. 

- X 

b 1 

o ^rmriTr^ 

"I II 

-glH|| 

o 

B0DI ^ 

lllliw 

V=%P, M=% PI, d= Pl 3 -f-48EI. 

V=K W, M=%wi, d=5Wl 3 -f-384E I. 

5 

L-to-- 


6 L J P 

i 

b 

A 

a>» a. 

b 


—^frrnrnTlTrr^ 

.Ik. 

c jtfm 

. 

V=Pa-^l, M= 

=Pab-^l. 

V=P, M=Pa, d=Pa (3l 2 -4a 2 )-f-24EI. 

7 ^ l P 

a ^ - i 


4 w-t 

miform load ^ 

A V _ ' 

' V* 

b 

1 

i 

b IIIlininT^— 

1 


:i: . i 

iifitw 

t > ... 

1 HlTTTTm>w. 

V=K P, M=%P1, d 

[=P1 3 -t-192EI. 

Y=KW,| M=AW1, 

v^n r 

d=Wl 3 -7-384EI. 














































































































































































































































































































































































































































































































































54 


STRENGTH OF MATERIALS 


parallel to a beam are called longitudinal forces; and others are 
called inclined forces. For the present we deal only with beams 
subjected to transverse forces (loads and reactions). 

57. Neutral Surface, Neutral Line, and Neutral Axis. When 
a beam is loaded it may be wholly convex up (concave down), as a 
cantilever; wholly convex down (concave up), as a simple beam 
on end supports; or partly convex up and partly convex down, as 
a simple beam with overhanging ends, a restrained beam, or a con. 



tinuous beam. Two vertical parallel lines drawn close together on 
the side of a beam before it is loaded will not be parallel after it 
is loaded and bent. If they are on a convex-down portion of a 
beam, they will be closer at the top and farther apart below than 
when drawn (Fig. 32 a), and if they are on a convex-up portion, 
they will be closer below and farther apart above than when drawn 
(Fig. 325). 

The “ fibres ” on the convex side of a beam are stretched and 
therefore under tension, while those on the concave side are short¬ 
ened and therefore under compression. Obviously there must be 
some intermediate fibres which are neither stretched nor shortened, 
i . e ., under neither tension nor compression. These make up 
a sheet of fibres and define a surface in the beam, which surface is 
called the neutral surface of the beam. The intersection of the 
neutral surface with either side of the beam is called the neutral 
line, and its intersection with any cross-section of the beam is 
called the neutral axis of that section. Thus, if ab is a fibre that 
has been neither lengthened nor shortened with the bending of the 
beam, then nn is a portion of the neutral line of the beam; and, 
if Fig. 32c be taken to represent a cross-section of the beam, NN 
is the neutral axis of the section. 

It can be proved that the neutral axis of any Gross-section of 











STRENGTH OF MATERIALS 


55 


a loaded beam passes through the center of gravity of that section , 
provided that all the forces applied to the beam are transverse, and 
that the tensile and compressive stresses at the cross-section are 
all within the elastic limit of the material of the beam. 

58. Kinds of Stress at a Cross=section of a Beam. It has 
already been explained in the preceding article that there are ten¬ 
sile and compressive stresses in a beam, and that the tensions are 
on the convex side of the beam and the compressions on the con¬ 
cave (see Fig. 33). The forces T and C are exerted upon the 
portion of the beam represented by the adjoining portion to the 



Fig. 33. 



right (not shown). These, the student is reminded, are often called 

fibre stresses. 

Besides the fibre stresses there is, in general, a shearing stress 
at every cross-section of a beam. This may be proved as follows: 

Fig. 34 represents a simple beam on end supports which has 
actually been cut into two parts as shown. The two parts can 
maintain loads when in a horizontal position, if forces are applied 
at the cut ends equivalent to the forces that would act there if the 
beam were not cut. Evidently in the solid beam there are at the 
section a compression above and a tension below, and such forces 
can be applied in the cut beam by means of a short block C and a 
chain or cord T, as shown. The block furnishes the compressive 
forces and the chain the tensile forces. At first sight it appears as 
if the beam would stand up under its load after the block and 
chain have been put into place. Except in certain cases*, how- 
ever, it would not remain in a horizontal position, as would the 


* When the external shear for the section is zero. 






















56 


STRENGTH OF MATERIALS 


solid beam. This shows that the forces exerted by the block and 
chain (horizontal compression and tension ) are not equivalent to 
the actual stresses in the solid beam. What is needed is a vertical 
force at each cut end. 

Suppose that R! is less than L x + L a + weight of A, i. e that 
the external shear for the section is negative; then, if vertical pulls 
be applied at the cut ends, upward on A and downward on B, the 
beam will stand under its load and in a horizontal position, just as 
a solid beam. These pulls can be supplied, in the model of the 
beam, by means of a cord S tied to two brackets fastened on A and 



Fig. 34. 


Fig. 35. 


B, as shown. In the solid beam the two parts act upon each 
other directly, and the vertical forces are shearing stresses, since 
they act in the plane of the surfaces to w T hich they are applied. 

59. Relation Between the Stress at a Section and the Loads 
and Reactions on Either Side of It. Let Fig. 35 represent the 
portion of a beam on the left of a section; and let R t denote the 
left reaction; Lj and L 2 the loads; W the weight of the left part; 

C, T, and S the compression, tension, and shear respectively which 
the right part exerts upon the left. 

Since the part of the beam here represented is at rest, all the 
forces exerted upon it are balanced; and when a number of hori¬ 
zontal and vertical forces are balanced, then 

1. The algebraic sum of the horizontal forces equals zero. 

2. “ “ “ “ “ vertical “ “ “ 

3. “ “ “ “ “ moments of all the forces with respect to 

any point equals zero. 

To satisfy condition 1, since the tension and compression are 
the only horizontal forces, the tension must equal the compression. 

To satisfy condition 2, S (the internal shear) must equal the 





















STRENGTH OF MATERIALS 


57 


algebraic sum of all the other vertical forces on the portion, that 
is, must equal the external shear for the section; also, S must act 
up or down according as the external shear is negative or positive. 
In other words, briefly expressed, the internal and external shears 
at a section are equal and opposite. 

To satisfy condition 3, the algebraic sum of the moments of 
the fibre stresses about the neutral axis must be equal to the sum 
of the moments of all the other forces acting on the portion about 
the same line, and the signs of those sums must be opposite. (The 
moment of the shear about the neutral axis is zero.) Now, the 
sum of the moments of the loads and reactions is called the bend¬ 
ing moment at the section, and if we use the term resisting mo= 
ment to signify the sum of the moments of the fibre stresses (ten¬ 
sions and compressions ) about the neutral axis, then we may say 
briefly that the resisting and the bending moments at a section are 
equal , and the two moments are opposite in sign. 

60. The Fibre Stress. As before stated, the fibre stress is 
not a uniform one, that is, it is not uniformly distributed over the 
section on which it acts. At any section, the compression is most 
“ intense ” (or the unit-compressive stress is greatest) on the con¬ 
cave side; the tension is most intense (or the unit-tensile stress is 
greatest) on the convex side; and the unit-compressive and unit- 
tensile stresses decrease toward the neutral axis, at which place the 
unit-fibre stress is zero. 

If the fibre stresses are within the elastic limit, then the two 
straight lines on the side of a beam referred to in Art. 57 will still 
be straight after the beam is bent; hence the elongations and short¬ 
enings of the fibres vary directly as their distance from the neutral 
axis. Since the stresses (if within the elastic limit) and deforma¬ 
tions in a given material are proportional, the unit-fibre stress 
varies as the distance from the neutral axis. 

Let Fig. 3Qa represent a portion of a bent beam, 36£ its cross- 
section, nn the neutral line, and NN the neutral axis. The way 
in which the unit-fibre stress on the section varies can be rep¬ 
resented graphically as follows: Lay off ac, by some scale, to 
represent the unit-fibre stress in the top fibre, and join c and n , 
extending the line to the lower side of the beam; also make be' equal 
to bc n and draw nc'. Then the arrows represent the unit-fibre 
stresses, for their ^ngthsvary as their distances from the neutral axis. 


58 


STRENGTH OF MATERIALS 


61. Value of the Resisting Moment. If S denotes the unit- 
fibre stress in the fibre farthest from the neutral axis (the greatest 
unit-fibre stress on the cross-section), and c the distance from the 
neutral axis to the remotest fibre, while S n S 2 , S 3 , etc., denote the 
unit-fibre stresses at points whose distances from the neutral axis 
are, respectively, y v y 2 , y 3 , etc. (see Fig. 3GJ), then 

S 

S : S, :: c : y x \ or Sj = — y v 


c 


Also, 
Let a 


j, (l^ #3, 


o s 0 s 

^2 n V'n ^3 2/3? e ^ c * 

o o 

etc., be the areas of the cross-sections of the fibres 



Fig. 36. 

whose distances from the neutral axis are, respectively, y v y 2 , y Zl 
etc. Then the stresses on those fibres are, respectively, 

S| ^15 S 3 ^25 S 3 ^35 etC. , 

s s s 


or. 


^ vft u q q etc. 


The arms of these forces or stresses wdth respect to the neutrai axis 
are, respectively, y n y 3 , etc.; hence their moments are 

s „ s _ s 


«i2/n 


^22/25 


0 ©tc., 


C 1 G 

and the sum of the moments (that is, the resisting moment) is 

s s s 

—a, y\ +—a 2 y\ + etc. = —(a l y\ 4- a 2 y\ + etc.) 

0 0 0 

Now y\ a 2 y\-\- etc. is the sum of the products obtained by 
multiplying each infinitesimal part of the area of the cross-section 
by the square of its distance from the neutral axis; hence, it is the 
moment of inertia of the cross-section with respect to the neutral 

axis. If this moment is denoted by I, then the value of the resist- 

, . SI 
ing moment is — 

o • 
















STRENGTH OF MATERIALS. 

PART II. 


STRENGTH OF BEAriS===(Concluded). 

62. First Beam Formula. As shown in the preceding 
article, the resisting and bending moments for any section of a 
beam are equal; hence 


SI T»f 

-7 = M > ( 6 ) 

all the symbols referring to the same section. This is the most 
important formula relating to beams, and will be called the “ first 
beam formula.” 

The ratio I -s- 0 is now quite generally called the section 
modulus. Observe that for a given beam it depends only on the 
dimensions of the cross-section, and not on the material or any¬ 
thing else. Since I is the product of four lengths (see article 51), 
I -r- c is the product of three; and hence a section modulus can be 
expressed in units of volume. The cubic inch is practically always 
used; and in this connection it is written thus, inches 3 . See Table 
A, page 52, for values of the section moduli of a few simple sections. 

63. Applications of the First Beam Formula. There are 
three principal applications of equation 6, which will now be ex¬ 
plained and illustrated. 

64. First Application. The dimensions of a beam and its 
manner of loading and support are given, and it is required to 
compute the greatest unit-tensile and compressive stresses in the 
beam. 

This problem can be solved by means of equation 6, w T ritten 
in this form, 


„ Me M 
S = "t or 


( 6 ') 


I ~ l-r-0 

Unless otherwise stated, we assume that the beams are uniform 
in cross-section, as they usually are; then the section modulus 
(I-^c) is the same for all sections, and S (the unit-fibre stress on 





60 


STRENGTH OF MATERIALS 


the remotest fibre) varies just as M varies, and is therefore greatest 
where M is a maximum.* Hence, to compute the value of the 
greatest unit-fibre stress in a given case, substitute the values of 
the section modulus and the maximum bending moment in the 
'preceding equation , and reduce. 

If the neutral axis is equally distant from the highest and low¬ 
est fibres, then the greatest tensile and compressive unit-stresses 
are equal, and their value is S. If the neutral axis is unequally 
distant from the highest and lowest fibres, let c denote its distance 
from the nearer of the tw T o, and S' the unit-fibre stress there. 
Then, since the unit-stresses in a cross-section are proportional to 
the distances from the neutral axis, 


S' 

s 


^ = —, or S' = —S. 


If the remotest fibre is on the convex side of the beam, S is tensile 
and S' compressive; if the remotest fibre is on the concave side, S 
is compressive and S' tensile. 

Examples. 1. A beam 10 feet long is supported at its ends, 
and sustains a load of 4,000 pounds two feet from the left end 
(Fig. 37, a'j. If the beam is 4 X 12 inches in cross-section (the 
long side vertical as usual), compute the maximum tensile and 
compressive unit-stresses. 

The section modulus of a rectangle whose base and altitude 
are b and a respectively (see Table A, page 52), is -\ba 2 \ hence, 
for the beam under consideration, the modulus is 

-rr~ X 4 X 12 2 = 96 inches 3 , 
u 


To compute the maximum bending moment, we have, first, to find 
the dangerous section. This section is where the shear changes 
sign (see article 45); hence, we have to construct the shear dia¬ 
gram, or as much thereof as is needed to find where the change of 
sign occurs. Therefore we need the values of the reaction. 
Neglecting the weight of the beam, the moment equation with 
origin at G (Fig. 37, a) is 

Rj X 10-4,000 X 8 = 0, or Rj = 3,200 pounds 

* Note. Because S is greatest in the section where M is maximum, this 
section is usually called the “ dangerous section ” of the beam. 





STRENGTH OF MATERIALS 


61 


Then, constructing the shear diagram, we see (Fig. 37, b) that the 
change of sign of the shear (also the dangerous section) is at the 
load. The value of the bending moment there is 

3,200 X 2 = 6,400 foot-pounds, 
or 6,400 X 12 = 76,800 inch-pounds. 

Substituting in equation 6 ', we find that 

S — —^ 75 — = 800 pounds per square men. 


U- 


fRT 


96 

4oooltos. 

£ l 1 


8 '- 


!B 


R 


i c (a) 



c’ (b) 


c" (c) 


Fig. 37. 

2» It is desired to take into account the weight of the beam 
in the preceding example, supposing the beam to be wooden. 

The volume of the beam is 

^ x 10 = 31 cubic feet; 

and supposing the timber to weigh 45 pounds per cubic foot, the 
beam weighs 150 pounds (insignificant compared to the load). 
The left reaction, therefore, is 

3,200 +(-g-X 150) = 3,275; 
















































































































































































































62 


STRENGTH OF MATERIALS 


and the shear diagram looks like Fig. 37, c, the shear changing 
sign at the load as before. The weight of the beam to the left of 
the dangerous section is 30 pounds; hence the maximum bending 
moment equals 


3,275 X 2 - 30 X 1 = 6,520 foot-pounds, 
6,520 X 12 = 78,240 inch-pounds. 


Substituting in equation 6 ', we find that 
q 78,240 

S = ■— 7 Tri — = 815 pounds pe 


The weight of the beam therefore increases the unit-stress pro¬ 
duced by the load at the dangerous section by 15 pounds per 
square inch. 

3. A T-bar (see Fig. 38) 8 feet long and supported at each 

end, bears a uniform load of 1,200 


c 


r 


o_ 



T 

c0 

CVJ 

<vi 



L 


Fig. 38. 


tral axis being 2.42 inches 4 , compute 
the maximum tensile and compressive 
unit-stresses in the beam 

Evidently the dangerous section 


is in the middle, and the value of the maximum bending moment 
(see Table B, page 53, Part I) is J Wl, W and l denoting the load 
and length respectively. Here 

— y?l = — x 1,200 X 8 = 1,200 foot-pounds, 


or 


1,200 X 12 = 14,400 inch-pounds. 

The section modulus equals 2.42 - 5 - 2.28 = 1.06; hence 
_ 14,400 

k — — q Q 0 - = 13,585 pounds per square inch. 


This is the unit-fibre stress on the lowest fibre at the middle sec¬ 
tion, and hence is tensile. On the highest fibre at the middle 
section the unit-stress is compressive, and equals (see page 60): 

c' 0.72 

^ ~ ~ g 28 ^ -^>^5 = 4,290 pounds per square inch. 













STRENGTH OF MATERIALS 


63 


Ans. 


EXAMPLES FOR PRACTICE. 

1. A beam 12 feet long and 6 X 12 inches in cross-section 
rests on end supports, and sustains a load of 3,000 pounds in the 
middle. Compute the greatest tensile and compressive unit- 
stresses in the beam, neglecting the weight of the beam. 

Ans. 750 pounds per square inch. 

2. Solve the preceding example taking into account the 
weight of the beam, 300 pounds 

Ans. 787.5 pounds per square inch. 

3. Suppose that a built-in cantilever projects 5 feet from the 
wall and sustains an end load of 250 pounds. The cross-section of 
the cantilever being represented in Fig. 38, compute the greatest 
tensile and compressive unit-stresses, and tell at what places they 
occur. (Neglect the weight.) 

Tensile, 4,471 pounds per square inch. 

Compressive, 14,150 “ “ “ “ 

4 . Compute the greatest tensile and compressive unit-stresses 
in the beam of Fig. 18, a, due to the loads and the weight of beam 
(400 pounds). (A moment diagram is represented in Fig. 18, l>\ 
for description see example 2, Art. 44, p. 39.) The section of 
the beam is a rectangle 8 X 12 inches. 

Ans. 580 pounds per square inch. 

5. Compute the greatest tensile and compressive unit-stresses 
in the cantilever beam of Fig. 19, a , it being a steel I-beam whose 
section modulus is 20.4 inches 3 . (A bending moment diagram for 
it is represented in Fig. 19, b; for description, see Ex. 3, Art. 44.) 

Ans. 11,470 pounds per square inch. 

6. Compute the greatest tensile and compressive unit-stresses 
in the beam of Fig. 10, neglecting its weight, the cross-sections 
being rectangular 6 X 12 inches. (See example for practice 1, 
Art. 43.) 

Ans. 600 pounds per square inch. 

65. Second Application. The dimensions and the work¬ 
ing strengths of a beam are given, and it is required to determine 
its safe load (the manner of application being given). 

This problem can be solved by means of equation 6 written 
in this form, 

M = — ( 6 ”) 


64 


STRENGTH OF MATERIALS 


We substitute for S the given working strength for the ma¬ 
terial of the beam, and for I and c their values as computed from 
the given dimensions of the cross-section; then reduce, thus 
obtaining the value of the safe resisting moment of the beam, 
which equals the greatest safe bending moment that the beam can 
stand. We next compute the value of the maximum bending 
moment in terms of the unknown load; equate this to the value 
of the resisting moment previously found; and solve for the 
unknown load. 

In cast iron, the tensile and compressive strengths are very 
different; and the smaller (the tensile) should always be used if 
the neutral surface of the beam is midway between the top and 
bottom of the beam; but if it is unequally distant from the top 
and bottom, proceed as in example 4, following. 

Examples. 1. A wooden beam 12 feet long and 6 X 12 
inches in cross-section rests on end supports. If its working 
strength is 800 pounds per square inch, how large a load uniformly 
distributed can it sustain ? 

The section modulus is \ba?, b and a denoting the base and 
altitude of the section (see Table A, page 52); and here 

77 ba 2 =^-rX 0 X 12 2 = 144 inches 3 , 
b b 


Hence S — = 800 X 144 = 115,200 inch-pounds. 

For a beam on end supports and sustaining a uniform load, the 
maximum bending moment equals JWZ (see Table B, page 55), 
W denoting the sum of the load and weight of beam, and l the 
length. If W is expressed in pounds, then 

‘w; = |w 


X 12 foot-pounds = W X 144 inch-pounds. 


Hence, equating the two values of maximum bending moment 
and the safe resisting moment, we get 


tt W X 144 = 115,200; 

o 

_ 115,200 X 8, 


jj-j--== 6,400 pounds. 


or, 


W 




STRENGTH OF MATERIALS 65 

The safe load for the beam is 6,400 pounds minus the weight of 
the beam. 

A steel I-beam whose section modulus is 20.4 inches 8 
rests on end supports 15 feet apart. Neglecting the weight of the 
beam, how large a load may be placed upon it 5 feet from one end, 
if the working strength is 16,000 pounds per square inch? 

The safe resisting moment is 

SI 

— — 16,000 X 20.4 = 326,400 inch-pounds; 

hence the bending moment must not exceed that value. The 
dangerous section is under the load; and if P denotes the unknown 
value of the load in pounds, the maximum moment (see Table B, 
page 53, Part I) equals § P X 5 foot-pounds, or § P X 60 inch- 
pounds. Equating values of bending and resisting moments, 
we get 

2 

■g P X 60 = 326,400; 

^ 326,400 X 3 

or > P = — 2 ~ — 0 q — = 8,160 pounds. 

3. In the preceding example, it is required to take into 
account the weight of the beam. 375 pounds. 




P 


I'- D 

> 

^ AW ^ 

t 


'h 


W= 375 llos. 

2k 


Fig. 39. 


As we do not know the value of the safe load, we cannot con¬ 
struct the shear diagram and thus determine where the dangerous 
section is. But in cases like this, where the distributed load (the 
weight) is small compared with the concentrated load, the dan¬ 
gerous section is practically always where it is under the concen¬ 
trated load alone; in this case, at the load. The reactions due to 
the weight equal X 375 = 187.5; and the reactions due to the 
load equal J P and § P, P denoting the value of the load. The 
larger reaction R 2 (Fig. 39) hence equals § P + 187.5. Since 













66 


STRENGTH OF MATERIALS 


the weight of the beam per foot is 375 -f- 15 — 25 pounds, the 
maximum bending moment (at the load) equals 

( | P + 187.5) 5 - (25 X 5) 2J = 

P p + 937.5 _ 312.5 = P P + 625. 

This is in foot-pounds if P is in pounds. 

The safe resisting moment is the same as in the preceding 
illustration, 326,400 inch-pounds; hence 

(4)’ P + 625) 12 = 326,400. 

Solving for P, we have 

10 P + 625 r= _ 826 ’ 400 - 


3 


12 


or. 


10 P + 625 X 3 = 

10 P = 79,725; 

P = 7,972.5 pounds. 


326,400 X 3 

12 


81,600; 


N 


It remains to test our assumption that the dangerous section 
is at the load. This can be done by computing R 1 (with I 5 = 
7,972.5), constructing the shear diagram, and noting where the 
shear changes sign. It will be found that the shear changes sign 
at the load, thus verifying the assumption. 

4. A cast-iron built-in cantilever beam projects 8 feet from 
the wall. Its cross-section is represented in Fig. 40, and the 

moment of inertia with respect to 
the neutral axis is 50 inches 4 ; the 
-jq working strengths in tension and 
compression are 2,000 and 9,000 
pounds per square inch respect¬ 
ively. Compute the safe uniform 
load which the beam can sustain, 
neglecting the weight of the beam. 

The beam being convex up, the upper fibres are in tension 
and the lower in compression. The resisting moment (SI -f- <?), 
as determined by the compressive strength, is 


~T 

in 

cvi 

1 

in 

4 -* 

JL 


Fig. 40. 























STRENGTH OF MATERIALS 


67 


9,000 X 50 
^L5 


100,000 inch-pounds; 


and the resisting moment, as determined by the tensile strength, is 

2,000 X 50 


2.5 


= 40.000 inch-pounds. 


Hence the safe resisting moment is the lesser of these two, or 
40,000 inch-pounds. The dangerous section is at the wall (see 
Table B, page 53), and the value of the maximum bending 
moment is Wl , W denoting the load and l the length. If W is 
in pounds, then 

M = -J W X 8 foot-pounds = | W X 96 inch-pounds. 
Equating bending and resisting moments, we have 

4-W X 96 = 40,000; 

bJ 


or, 


W 


40,000 X 2 
96 


= 833 pounds. 


EXAMPLES FOR PRACTICE. 

1. An 8 X 8 -inch timber projects 8 feet from a wall. If its 
working strength is 1,000 pounds per square inch, how large an 
end load can it safely sustain ? 

Ans. 890 pounds. 

2. A beam 12 feet long and 8 X 16 inches in cross-section, 
on end supports, sustains two loads P, each 3 feet from its ends 
respectively. The working strength being 1,000 pounds per square 
inch, compute P (see Table B, page 53). 

Ans. 9,480 pounds. 

3. An I-beam weighing 25 pounds per foot rests on end 
supports 20 feet apart. Its section modulus is 20.4 inches 3 , and 
its working strength 16,000 pounds per square inch. Compute 
the safe uniform load which it can sustain. 

Ans. 10,880 pounds- 

66. Third Application. The loads, manner of support, 
and working strength of beam are given, and it is required to de¬ 
termine the size of cross-section necessary to sustain the load 
safely, that is, to “design the beam.” 






68 


STRENGTH OF MATERIALS 


To solve this problem, we use the first beam formula (equation 
6 ), written in this form, 

_I _M. (6'") 

c “ S 


We first determine the maximum bending moment, and then sub¬ 
stitute its value for M, and the working strength for S. Then we 
have the value of the section modulus (I - 4 - c) of the required 
beam. Many cross-sections can be designed, all having a given 
section modulus. Which one is to be selected as most suitable will 
depend on the circumstances attending the use of the beam and 
on considerations of economy. 

Examples. 1. A timber beam is to be used for sustaining 
a uniform load of 1,500 pounds, the distance between the supports 
being 20 feet. If the working strength of the timber is 1,000 pounds 
per square inch, what is the necessary size of cross-section ? 

The dangerous section is at the middle of the beam; and the 
maximum bending moment (see Table B, page 53) is 


— WZ = -g- x 1,500 X 20 = 3,750 foot-pounds, 


or 


Hence 


3,750 X 12 = 45,000 inch-pounds. 
I 45.000 ; 4g incheg3 _ 


c 1,000 

Now the section modulus of a rectangle is \ba? (see Table A, 
page 54, Part I); therefore, \ba 2 — 45, or ba 2 = 270. 

Any wooden beam (safe strength 1,000 pounds per square 
inch) whose breadth times its depth square equals or exceeds 270, 
is strong enough to sustain the load specified, 1,500 pounds. 

To determine a size, we may choose any value for b or a, and 
solve the last equation for the unknown dimension. It is best, 
however, to select a value of the breadth, as 1, 2, 3, or 4 inches, 
and solve for a. Thus, if we try b = 1 inch, we have 


a 2 = 270, or a = 16.43 inches. 

This would mean a board 1 X 18 inches, which, if used, would 
have to be supported sidewise so as to prevent it from tipping or 
“ buckling.” Ordinarily, this would not be a good size. 

Next try b — 2 inches; we have 

2 X a 2 = 270; or a — i/270 - 4 - 2 = 11.62 inches. 

This would require a plank 2 X 12, a better proportion than the 
first. Trying b = 3 inches, we have 







STRENGTH OF MATERIALS 


69 


3 X ci 2 = 270; or a — V 270 h- 3 = 0.49 inches. 

This would require a plank 3 X 10 inches; and a choice between 
a 2 X 12 and a 3 X 10 plank would be governed by circumstances 
in the case of an actual construction. 

It will be noticed that we have neglected the weight of the 
beam. Since the dimensions of wooden beams are not fractional, 
and we have to select a commercial size next larger than the one 
computed (12 inches instead of 11.62 inches, for example), the 
additional depth is usually sufficient to provide strength for the 
weight of the beam. If there is any doubt in the matter, we can 
settle it by computing the maximum bending moment including 
the weight of the beam, and then computing the greatest unit-fibre 
stress due to load and weight. If this is less than the safe strength, 
the section is large enough; if greater, the section is too small. 

Thus, let us determine whether the 2 X 12-inch plank is 
strong enough to sustain the load and its own weight. The plank 
will weigh about 120 pounds, making a total load of 

1,500 -j- 120 = 1,620 pounds. 

Hence the maximum bending moment is 

— -—1,620 X 20 X 12 = 48,600 inch-pounds. 

o o 

Since -1 = A Id? = Ax 2 X 12 2 = 48, and S = J—, 

G o b 1 -r-c* 

48,600 . 

IS =- -rr~- = 1,013 pounds per square inch. 


Strictly, therefore, the 2 X 12-inch plank is not large enough; but 
as the greatest unit-stress in it would be only 13 pounds per square 
inch too large, its use would be permissible. 

2. What size of steel I-beam is needed to sustain safely the 
loading of Fig. 9 if the safe strength of the steel is 16,000 pounds 
per square inch ? 

The maximum bending moment due to the loads was found 
in example 1, Art. 43, to be 8,800 foot-pounds, or 8,800 X 12 = 
105,600 inch-pounds. 


Hence 



105,600 

16,000 


= 6.6 inches 3 . 


That is, an I-beam is needed whose section modulus is a little 
larger than 6.6, to provide strength for its own weight. 









70 


STRENGTH OF MATERIALS 


To select a size, we need a descriptive table of I-beams, such 
as is published in handbooks on structural steel. 

Below is an abridged copy of such a table. (The last two columns con¬ 
tain information for use later.) The figure illustrates a cross-section of an 
I-beam, and shows the axes referred to in the table. 

It will be noticed that two sizes are given for each depth; 
these are the lightest and heaviest of each size that are made, but 
intermediate sizes can be secured. In column 5 we find 7.3 as the 
next larger section modulus than the one required (6,6); and this 
corresponds to a 12^-pound 6-inch I-beam, which is probably the 
proper size. To ascertain whether the excess (7.3-6.6 = 0.70) 
in the section modulus is sufficient to provide for the weight of the 
beam, we might proceed as in example 1. In this case, however, 
the excess is quite large, and the beam selected is doubtless safe. 

TABLE C. 

Properties ot Standard I*Beams 



[L. 5 J\ 



\l r \ 



Section of beam, showing axes 1-1 and 2-2. 


1 

2 

3 

4 

5 

6 

Depth of 

Weight 

Area of cross- 

Moment of 

Section 

Moment of 

Beam, 

per foot, 

section,in 

inertia. 

modulus, 

inertia, 

in inches. 

in pounds. 

square inches. 

axis 1—1. 

axis 1—1. 

axis 2—2. 

3 

5.50 

1.63 

2.5 

1.7 

0.46 

3 

7.50 

2.21 

2.9 

1.9 

.60 

4 

7.50 

2.21 

6.0 

3.0 

.77 

4 

10.50 

3.09 

7.1 

3.6 

1.01 

5 

9.75 

2.87 

12.1 

4.8 

1.23 

5 

14.75 

4.34 

15.1 

6.1 

1.70 

6 

12.25 

3.61 

21.8 

7.3 

1.85 

6 

17.25 

5.07 

26.2 

8.7 

2.36 

7 

15.00 

4.42 

36.2 

10.4 

2.67 

7 

20.00 

5.88 

42.2 

12.1 

3.24 

8 

18.00 

5.33 

56.9 

14.2 

3.78 

8 

25.25 

7.43 

68.0 

17.0 

4.71 

9 

21.00 

6.31 

84.9 

18.9 

5.16 

9 

35.00 

10.29 

111.8 

24.8 

7.31 

10 

25.00 

7.37 

122.1 

24.4 

6.89 

10 

40.00 

11.76 

158.7 

31.7 

9.50 

12 

31.50 

9.26 

215.8 

36.0 

9.50 

12 

40.00 

11.76 

245.9 

41.0 

10.95 

15 

42.00 

12.48 

441.8 

58.9 

14.62 

15 

60.00 

17.65 

538.6 

71.8 

18.17 

18 

55.00 

15.93 

795.6 

88.4 

21.19 

18 

70.00 

20.59 

921.2 

102.4 

24.62 

20 

65.00 

19.08 

1 , 169.5 

117.0 

27.86 

20 

75.00 

22.06 

1 , 268.8 

126.9 

30.25 

24 

80.00 

23.32 

2 , 087.2 

173.9 

42.86 

24 

100.00 

29.41 

2 , 379.6 

198.3 

48.55 




























STRENGTH OF MATERIALS 


71 


EXAMPLES FOR PRACTICE. 

1. Determine the size of a wooden beam which can safely 
sustain a middle load of 2,000 pounds, if the beam rests on end 
supports 16 feet apart, and its working strength is 1,000 pounds 
per square inch. Assume width 6 inches. 

Ans. 6 X 10 inches. 

2. What sized steel I-beam is needed to sustain safely a 
uniform load of 200,000 pounds, if it rests on end supports 10 
feet apart, and its working strength is 16,000 pounds per square 

inch? 

Ans. 95-pound 24-inch. 

3. What sized steel I-beam is needed to sustain safely the 
loading of Fig. 10, if its working strength is 16,000 pounds per 
square inch ? 

Ans. 14.75-pound 5-inch. 

67. Laws of Strength of Beams. The strength of a beam is 
measured by the bending moment that it can safely withstand; or, 
since bending and resisting moments are equal, by its safe resist, 
ing moment (SI -r- c). Hence the safe strength of a beam varies 
(1) directly as the working fibre strength of its material, and (2) 
directly as the section modulus of its cross-section. For beams 
rectangular in cross-section (as wooden beams), the section modu¬ 
lus is \bd\ b and a denoting the breadth and altitude of the 
rectangle. Hence the strength of such beams varies also directly 
as the breadth, and as the square of the depth. Thus, doubling 
the breadth of the section for a rectangular beam doubles the 
strength, but doubling the depth quadruples the strength. 

The safe load that a beam can sustain varies directly as its 
resisting moment, and depends on the way in which the load is 
distributed and how the beam is supported. Thus, in the first 
four and last two cases of the table on page 55, 


M = FI, hence 
M = i W l, “ 

M = J FI, “ 

M = -i W l, “ 

M = | FI, “ 

M = T V W l, “ 


P = SI -s- lo, 
W = 2SI -h lo, 
P = 4SI -h lo, 
W = 8SI -4- lo, 
P = 8SI -4- lo, 
W = 12SI h- lo. 


72 


STRENGTH OF MATERIALS 


Therefore the safe load in all cases varies inversely with the 
length; and for the different cases the safe loads are as 1, 2, 4, 8, 
8, and 12 respectively. 

Example. What is the ratio of the strengths of a plank 2 X 
10 inches when placed edgewise and when placed flatwise on its 
supports ? 

When placed edgewise, the section modulus of the plank is 
-i X 2 X 10 2 = 33J, and when placed flatwise it is -J X 10 X 2 2 — 
6|; hence its strengths in the two positions are as 33J to 6| 
respectively, or as 5 to 1. 

EXAMPLE FOR PRACTICE. 

What is the ratio of the safe loads for two beams of wood, 
one being 10 feet long, 3 X 12 inches in section, and having its load 
in the middle; and the other 8 feet long and 2x8 inches in section, 
with its load uniformly distributed. 

Ans. As 28.8 to 21.3 

68. Modulus of Rupture. If a beam is loaded to destruction, 
and the value of the bending moment for the rupture stage is 
computed and substituted for M in the formula SI -s- c = M, then 
the value of S computed from the equation is the modulus of 
rupture for the material of the beam. Many experiments have 
been performed to ascertain the moduli of rupture for different 
materials and for different grades of the same material. The fol- 
owing are fair values, all in pounds per square inch: 

TABLE D. 

Moduli of Rupture. 

Timber: 

Spruce. 

Hemlock. 

White pine. 

Long-leaf pine.... 

Short-leaf pine... 

Douglas spruce... 

White oak. 

Red oak. 


Stone: 

Sandstone.. 

400— 1 200 


Limestone. 

400 1,000. 

800 1,400. 


Granite. 


Cast iron: 

One and one-half to two 

one-quarter times its 
mate tensile strength. 

and 

ulti- 

Hard steel: 

Varies from 100,000 to 150,000 


4,000- 

- 7,000, 

average 

5,000 

3,500 

7,000, 

u 

4,500 

5,500 

10,500, 

it 

8,000 

10,000 

16,000, 

<• 

12,500 

8,000 

14,000, 

u 

10,000 

4,000 

12,000, 

u 

8,000 

7,500 

18,500, 

u 

13,000 

9,000 

15,000, 

u 

11,500 




















STRENGTH OF MATERIALS 


73 


Wrought iron and structural steels have no modulus of rup¬ 
ture, as specimens of those materials will “ bend double,” but not 
break. The modulus of rupture of a material is used principally 
as a basis for determining its working strength. The factor of 
safety of a loaded beam is computed by dividing the modulus 
of rupture of its maternal by the greatest unit-fibre stress in 
the beam . 

69. The Resisting Shear. The shearing stress on a cross- 
section of a loaded beam is not a uniform stress; that is, it is not 
uniformly distributed over the section. In fact the intensity or 
unit-stress is actually zero on the highest and lowest fibres of a 
cross-section, and is greatest, in such beams as are used in prac¬ 
tice, on fibres at the neutral axis. In the following article we 
explain howto find the maximum value in two cases—cases which 
are practically important. 

70. Second Beam Formula. Let S s denote the average 
value of the unit-shearing stress on a cross-section of a loaded 
beam, and A the area of the cross-section. Then the value of the 
whole shearing stress on the section is : 

Resisting shear = S G A. 

O S 

Since the resisting shear and the external shear at any section of a 
beam are equal (see Art. 59), 

S s A = V. (7) 

This is called the “ second beam formula ” It is used to investi¬ 
gate and to design for shear in beams. 

In beams uniform in cross-section, A is constant, and S s is 
greatest in the section for which V is greatest. Hence the great¬ 
est unit-shearing stress in a loaded beam is at the neutral axis of 
the section at w T hich the external shear is a maximum. There is 
a formula for computing this maximum value in any case, but it 
is not simple, and we give a simpler method for computing the 
value in the two practically important cases: 

1. In wooden beams (rectangular or square in cross-section), the 
greatest unit-shearing stress in a section is 50 per cent larger than the average 

value S g . 

2. In I-beams, and in others with a thin vertical web, the greatest 
unit-shearing stress in a section practically equals S 8 , as given by equation 7, 
if the area of the web is substituted for A. 


74 


STRENGTH OF MATERIALS 


Examples. 1 . What is the greatest value of the unit¬ 
shearing stress in a wooden beam 12 feet long and 6 X 12 inches in 
cross-section when resting on end supports and sustaining a uni¬ 
form load of 6,400 pounds ? (This is the safe load as determined 
by working fibre stress; see example 1, Art. 65.) 

The maximum external shear equals one-half the load (see 
Table B, page 53), and comes on the sections near the supports. 


Since 


A = 6 X 12 = 72 square inches; 


3,200 

~72~ 


44 pounds per square inch, 


and the greatest unit-shearing stress equals 
3 3 

~ 2 ~ S s = -g- 44 = 66 pounds per square inch. 

Apparently this is very insignificant; but it is not negligible, as 
is explained in the next article. 

2. A steel I-beam resting on end supports 15 feet apart 
sustains a load of 8,000 pounds 5 feet from one end. The weight 
of the beam is 375 pounds, and the area of its web section is 3.2 
square inches. (This is the beam and load described in examples 
2 and 3, Art. 65.) What is the greatest unit-shearing stress ? 

The maximum external shear occurs near the support where 
the reaction is the greater, and its value equals that reaction. 
Calling that reaction R, and taking moments about the other end 
of the beam, we have 

R X 15 — 375 x 7y - 8,000 X 10 = 0; 

therefore 15 R = 80,000 -f- 2,812.5 = 82,812.5; 
or, R = 5,520.8 pounds. 

„ G 5,520.8 . 

Hence b 8 ==— ^ ^ • = 1,725 pounds per square inch. 


EXAMPLES FOR PRACTICE. 

1. A wooden beam 10 feet long and 2 X 10 inches in cross- 
section sustains a middle load of 1,000 pounds. Neglecting the 
weight of the beam, compute the value of the greatest unit-shearing 
stress. 

Ans. 37.5 pounds per square inch. 






STRENGTH OF MATERIALS 


75 


2. Solve the preceding example taking into account the 
weight of the beam, 60 pounds. 

Ans. 40 pounds per square inch. 

3. A wooden beam 12 feet long and 4 X 12 inches in cross- 
section sustains a load of 3,000 pounds 4 feet from one end. 
Neglecting the weight of the beam, compute the value of the 
greatest shearing unit-stress. 

Ans. 62.5 pounds per square inch. 

71. Horizontal Shear. It can be proved that there is a 
shearing stress on every horizontal section of a loaded beam. An 
experimental explanation will have to suffice here. Imagine a 
pile of six boards of equal length supported so that they do not 
bend. If the intermediate supports are removed, they will bend 
and their ends will not be flush but somewhat as represented in 
Fig. 41. This indicates that the boards slid over each other during 
the bending, and hence there was a rubbing and a frictional re¬ 
sistance exerted by the boards upon each other. Now, when a 
solid beam is being bent, there is an exactly similar tendency for 
the horizontal layers to slide over each other; and, instead of a 
frictional resistance, there exists shearing stress on all horizontal 
sections of the beam. 

In the pile of boards the amount of slipping is different at 
different places between any two boards, being greatest near the 
supports and zero midway between them. Also, in any cross- 
section the slippage is least between the upper two and lower two 
hoards, and is greatest between the middle two. These facts indi¬ 
cate that the shearing unit-stress on horizontal sections of a solid 

O 

beam is greatest in the neutral surface at the supports. 

It can be proved that at any place in a heam the shearing 
unit-stresses on a horizontal and on a vertical section are equal. 



Fig. 41. Fig. 42. 

It follows that the horizontal shearing unit-stress is greatest at the 
neutral axis of the section for which the external shear (Y) is a 
maximum. Wood being very weak in shear along the grain, 
timber beams sometimes fail under shear, the “ rupture ” being 






76 


STRENGTH OF MATERIALS 


two horizontal cracks along the neutral surface somewhat as rep¬ 
resented in Fig. 42. It is therefore necessary, when dealing with 
timber beams, to give due attention to their strength as determined 
by the working strength of the material in shear along the grain. 

Example. A wooden beam 3 X 10 inches in cross-section 
rests on end supports and sustains a uniform load of 4,000 pounds 
Compute the greatest horizontal unit-stress in the beam. 

The maximum shear equals one-half the load (see Table B, 
page 55), or 2,000 pounds. Hence, by equation 7, since A = 
3 X 10 = 30 square inches, 


2,000 


66-q- pounds per square inch. 


This is the average shearing unit-stress on the cross-sections near 
the supports; and the greatest value equals 

3 2 

X 66 — == 100 pounds per square inch. 


According to the foregoing, this is also the value of the 
greatest horizontal shearing unit-stress. (If of white pine, for 
example, the beam would not be regarded as safe, since the ulti¬ 
mate shearing strength along the grain of selected pine is only 
about 400 pounds per square inch.) 

72. Design of Timber Beams. In any case we may pro¬ 
ceed as follows:—(1) Determine the dimensions of the cross- 
section of the beam from a consideration of the fibre stresses as 
explained in Art. 66 . (2) With dimensions thus determined, com¬ 
pute the value of the greatest shearing unit-stress from the formula, 

3 

Greatest shearing unit-stress = Y -r- ab , 


where Y denotes the maximum external shear in the beam, and 
b and a the breadth and depth of the cross-section. 

If the value of the greatest shearing unit-stress so computed 
does not exceed the working strength in shear along the grain, 
then the dimensions are large enough; but if it exceeds that value, 
then a or b, or both, should be increased until | Y -r- ab is less 
than the working strength. Because timber beams are very often 
“season checked” (cracked) along the neutral surface, it is advis- 



STRENGTH OF MATERIALS 


77 


able to take the working strength of wooden beams, in shear along 
the grain, (juite low. One-twentieth of the working fibre strength 
has been recommended* for all pine beams. 

If the working strength in shear is taken equal to one- 
twentieth the working fibre strength, then it can be shown that, 

1. For a beam on end supports loaded in the middle, the safe load de¬ 
pends on the shearing or fibre strength according as the ratio of length to 
depth (l + a) is less or greater than 10. 

2. For a beam on end supports uniformly loaded, the safe load depends 
on the shearing or fibre strength according as l a is less or greater than 20. 

Examples. 1. It is required to design a timber beam to sus¬ 
tain loads as represented in Fig. 11, the working fibre strength 
being 550 pounds and the working shearing strength 50 pounds 
per square inch. 

The maximum bending moment (see example for practice 3, 
Art. 43; and example for practice 2, Art. 44) equals practically 
7,000 foot-pounds or, 7,000 X 12 = 84,000 inch-pounds. 

Hence, according to equation 6'", 


I 


c 


84,000 

”66<r 


= 152.7 inches 8 . 


Since for a rectangle 



-I ba? = 152.7, or ba 2 = 916.2. 

Now, if we let l> — 4, then a 2 = 229; 

or, a — 15.1 (practically 16) inches. 

If, again, we let b — 6, then a 2 — 152.7; 

or a — 12.4 (practically 1*4) inches. 

Either of these sizes will answer so far as fibre stress is concerned, 
but there is more u timber” in the second. 

The maximum external shear in the beam equals 1,556 
pounds, neglecting the weight of the beam (see example 3, Art. 
37; and example 2, Art. 38). Therefore, for a 4 X 16-inch beam, 


* See “ Materials of Construction.”— Johnson. Page 55. 






78 


STRENGTH OF MATERIALS 


Greatest shearing unit-stress = 



1,556 
4 X 16 


= 36.5 pounds per square inch; 


and for a 6 X 14-inch beam, it equals 


3 1,556 

if X 6 X 14 


27.7 pounds per square inch. 


Since these values are less than the working strength in shear, 
either size of beam is safe as regards shear. 

o 

If it is desired to allow for weight of beam, one of the sizes 
should be selected. First, its weight should be computed, then 
the new reactions, and then the unit-fibre stress may be com¬ 
puted as in Art. 64, and the greatest shearing unit-stress as in the 
foregoing. If these values are within the working values, then 
the size is large enough to sustain safely the load and the weight 
of the beam. 

2. What is the safe load for a white pine beam 9 feet long 
and 2x12 inches in cross-section, if the beam rests on end supports 
and the load is at the middle of the beam, the working fibre 
strength being 1,000 pounds and the shearing strength 50 pounds 
per square inch. 

The ratio of the length to the depth is less than 10; hence 
the safe load depends on the shearing strength of the material 
Calling the load P, the maximum external shear (see Table B, 
page 53) equals i P, and the formula for greatest shearing unit 
stress becomes 

3 1 p 

50 = ~ 2 ~X or P = 1,600 pounds. 


EXAMPLES FOR PRACTICE. 

1. What size of wooden beam can safely sustain loads as in 
Fig. 12, with shearing and fibre working strength equal to 50 and 
1,000 pounds per square inch respectively ? 

Ans. 6 X 12 inches 

2. What is the safe load for a wooden beam 4 X 14 inches, 
and 18 feet long, if the beam rests on end supports and the load 
is uniformly distributed, with working strengths as in example 1 ? 

Ans. 3,730 pounds 








STRENGTH OF MATERIALS 


79 


73. Kinds of Loads and Beams. We shall now discuss the 
strength of beams under longitudinal forces (acting parallel to 
the beam) and transverse loads. The longitudinal forces are 
supposed to be applied at the ends of the beams and along the axis* 
of the beam in each case. We consider only beams resting on 
end supports. 

The transverse forces produce bending or flexure, and the 
longitudinal or end forces, if pulls, produce tension in the beam; 
if pushes, they produce compression. Hence the cases to be con¬ 
sidered may be called “ Combined Flexure and Tension ” and 
“ Combined Flexure and Compression.” 

74. Flexure and Tension. Let Fig. 43, a , represent a beam 
subjected to the transverse loads L l5 L 2 and L 3 , and to two equal 
end pulls P and P. The reactions Rj and R 2 are due to the trans¬ 
verse loads and can be computed by the methods of moments just 
as though there were no end pulls. To find the stresses at any 
cross-section, we determine those due to the transverse forces 
(Lj, L 2 , L 3 , Rj and R 2 ) and those due to the longitudinal; then 
combine these stresses to get the total effect of all the applied 
forces. 

The stress due to the transverse forces consists of a shearing 
stress and a fibre stress; it will be called the flexural stress. The 
fibre stress is compressive above and tensile below. Let M denote 
the value of the bending moment at the section considered; c, and 
c 2 the distances from the neutral axis to the highest and the low¬ 
est fibre in the section; and Sj and S 2 the corresponding unit-fibre 
stresses due to the transverse loads. Then 



Mg, 


; and S 2 = 


Me, 

I ' 


The stress due to the end pulls is a simple tension, and it equals 
P; this is sometimes called the direct stress. Let S Q denote the 
unit-tension due to P, and A the area of the cross-section; then 



A* 


Both systems of loads to the left of a section between L } and 

* Note. By “ axis of a beam ” is meant the line through the centers of 
gravity of all the cross-sections. 





80 


STRENGTH OF MATERIALS 


L 2 are represented in Fig. 48, Z»; also the stresses caused by them 
at that section. Clearly the effect of the end pulls is to increase the 


tensile stress (on the lower 
fibres) and to decrease the 
compressive stress (on the 
upper fibres) due to the flex¬ 
ure. Let S c denote the total 
(resultant) unit-stress on the 
upper fibre, and S t that on 
the lower fibre, due to all 
the forces acting on the beam. 
In combining the stresses 
there are two cases to con¬ 
sider: 



Fig. 43. 


(1) The flexural compressive unit-stress on the upper fibre is 
greater than the direct unit-stress; that is, S 2 is greater than S Q . 
The resultant stress on the upper fibre is 

S c = Sj — S G (compressive); 
and that on the lower fibre is 

S t = S 2 + S G (tensile). 

The combined stress is as represented in Fig. 43, <?, part tensile 
and part compressive. 

(2) The flexural compressive unit-stress is less than the 
direct unit-stress; that is, S, is less than S Q . Then the combined 
unit-stress on the upper fibre is 

S c = S G - S, (tensile); 
and that on the lower fibre is 

S t = S 2 + S G (tensile). 

The combined stress is represented by Fig. 43, d , and is all 
tensile. 

Example. A steel bar 2x6 inches, and 12 feet long, is sub¬ 
jected to end pulls of 45,000 pounds. It is supported at each 
end, and sustains, as a beam, a uniform load of 6,000 pounds. 
It is required to compute the combined unit-fibre stresses. 

Evidently the dangerous section is at the middle, and M = 
■§- WZ; that is, 
























STRENGTH OF MATERIALS 


81 



X 6,000 X 12 = 9,000 foot-pounds, 


or 9,000 X 12 = 108,000 inch-pounds. 

The bar being placed with the six-inch side vertical, 
c x = c 2 = 3 inches, and 

I = i x 2 X 6 3 = 36 inches 4 . (See Art. 52.) 


Hence 

Since 


s i = s 2 


108,000 X 3 
36 


9,000 pounds per square inch. 


A = 2 X 6 = 12 square inches, 

S 0 = = *V^0 pounds per square inch. 


The greatest value of the combined compressive stress is 

Sj - S Q = 9,000 - 3,750 = 5,250 pounds per square inch, 
and it occurs on the upper fibres of the middle section. The great¬ 
est value of the combined tensile stress is 

S 2 4~ S Q = 9,000 -f- 3,750 = 12,750 pounds per square inch, 
and it occurs on the lowest fibres of the middle section. 


EXAHPLE FOR PRACTICE. 


Change the load in the preceding illustration to one of 6,000 
pounds placed in the middle, and then solve. 

* j S c = 14,250 pounds per square inch. 
Ans * } S t = 21,750 “ 


75. Flexure and Compression. Imagine the arrowheads on 
F reversed; then Fig. 43, a, will represent a beam under com¬ 
bined flexural and compressive stresses. The flexural unit-stresses 
are computed as in the preceding article. The direct stress is a 
compression equal to P, and the unit-stress due to P is computed 
as in the preceding article. Evidently the effect of P is to increase 
the compressive stress and decrease the tensile stress due to the 
flexure. In combining, we have two cases as before: 

(1) The flexural tensile unit-stress is greater than the 
direct unit-stress; that is, S 2 is greater than S Q . Then the com¬ 
bined unit-stress on the lower fibre is 






82 


STRENGTH OF MATERIALS 


S t = S 2 - S 0 (tensile); 
and that on the upper fibre is 

— S, + S G (compressive).. 

The combined fibre stress is represented by Fig. 44, a , and is part 
tensile and part compressive. 

(2) The flexural unit-stress on the lower fibre is less than 
the direct unit-stress; that is, S 2 is less than S G . Then the com¬ 
bined unit-stress on the lower fibre is 
S t = S Q - S 2 (compressive); 
and that on the upper fibre is 
S c = S G + (compressive). 

The combined fibre stress is represented by 
Fig. 44, b , and is all compressive. 

Example. A piece of timber 6x6 
inches, and 10 feet long, is subjected to end 
pushes of 9,000 pounds. It is supported in 
a horizontal position at its ends, and sustains 
a middle load of 400 pounds. Compute the 
combined fibre stresses. 

Evidently the dangerous section is at the 
middle, and M = \ P l\ that is, 

M = | X 400 X 10 = 1,000 foot-pounds, 



Fig. 44. 


or 

Since 


Since 


1,000 X 12 = 12,000 inch-pounds. 
c x — c 2 = 3 inches, and 

I = -jg ba 3 = X 6 X 6 3 = 108 inches 4 , 

12,000 X 3 1 

ki = - = 333-^- pounds per square inch, 

A = 6 X 6 = 36 square inches, 

9,000 


S 


o 


36 


250 pounds per square inch. 


Hence the greatest value of the combined compressive stress is 
So + Sj = 333-g- -f- 250 = 583-g- pounds per square inch. 





























STRENGTH OF MATERIALS 


83 


It occurs on the upper fibres of the middle section. The greatest 
value of the combined tensile stress is 


S 2 - S Q = 333-g- _ 250 = 83— pounds per square inch. 
It occurs on the lowest fibres of the middle section. 


EXAMPLE FOR PRACTICE. 

Change the load of the preceding illustration to a uniform 
load and solve. 

Ans j Sc = 417 pounds per square inch. 

( S t = 83 “ “ “ “ (compression). 

76. Combined Flexural and Direct Stress by flore Exact 
Formulas. The results in the preceding articles are only approxi¬ 
mately correct. Imagine the 
beam represented in Fig. 45, a , 
to be first loaded with the trans¬ 
verse loads alone. They cause 
the beam to bend more or less, 
and produce certain flexural 
stresses at each section of the 
beam. Now, if end pulls are 
applied they tend to straighten 
the beam and hence diminish the flexural stresses. This effect 
of the end pulls was omitted in the discussion of Art. 74, and 
the results there given are therefore only approximate, the 
value of the greatest combined fibre unit-stress (S t ) being too 
large. On the other hand, if the end forces are pushes, they in¬ 
crease the bending, and therefore increase the flexural fibre stresses 
already caused by the transverse forces (see Fig. 45, b). The 
results indicated in Art. 75 must therefore in this case also be 
regarded as only approximate, the value of the greatest unit- 
fibre stress (S c ) being too small. 

For beams loaded in the middle or with a uniform load, the 
following formulas, which take into account the flexural effect of 
the end forces, may be used: 

M denotes bending moment at the middle section of the beam; 

I denotes the moment of inertia of the middle section with 
respect to the neutral axis; 



b 

Fig. 45. 













84 


STRENGTH OF MATERIALS 


Sj, S 2 , c x and c 2 have the same meanings as in Arts. 74 and 
75, but refer always to the middle section; 
l denotes length of the beam; 

E is a number depending on the stiffness of the material, the 
average values of which are, for timber, 1,500,000; and for struc¬ 
tural steel 30,000,000.* 



M o x 

yqrpF 

10E 


and S, 


M * 2 

! + W ' ( 8 ) 

10E 


The plus sign is to be used when the end forces P are pulls, and 
the minus sign when they are pushes. 

It must be remembered that S t and S 2 are flexural unit- 
stresses. The combination of these and the direct unit-stress is 
made exactly as in articles 74 and 75. 

Examples. 1. It is required to apply the formulas of this 
article to the example of article 74. 

As explained in the example referred to, M = 108,000 inch- 
pounds; c = c 2 = 3 inches; and I = 36 inches 4 . 

Now, since l — 12 feet = 144 inches, 

_ _ 108,000 X 3 

1 “ 2 ” 45,000 X 14P~ 

+ 10 X 30,000,000 

per square inch, as compared with 9,000 pounds per square inch, 
the result reached by the use of the approximate formula. 

As before, S 0 = 3,750 pounds per square inch; hence 

S c = 8,284- 3,750 = 4,534 pounds per square inch; 
and S t = 8,284 + 3,750 = 12,034 “ “ “ “ 

2. It is required to apply the formulas of this article to the 
example of article 75. 

As explained in that example, 

M — 12,000 inch-pounds; 
c x = c 2 =. 3 inches, and I = 108 inches 4 . 

Now, since l = 120 inches, 

Q « _ 12,000 X 3 36,000 

1 2 9,000 x 120 2 — 108 - 8.64 — 302 P ounds 

' "10 X 1,600,000 


324,000 n no , 

36 | 3 11 8,^84 pounds 


* Note. This quantity “ E ” is more fully explained in Article 95. 





















STRENGTH OF MATERIALS 


85 


pei square incli, as compared with 333^ pounds per square inch, 
the result reached by use of the approximate method. 

As before, S 0 = 250 pounds per square inch; hence 

S c = 362 + 250 = 612 pounds per square inch; and 
S t = 362 - 250 = 112 « « « « 


EXAMPLES FOR PRACTICE. 

1* Solve the example for practice of Art. 74 by the formulas 
of this article. 

Ans ) = 12,820 pounds per square inch. 


20,320 “ 


u 


2. Solve the example for practice of Art. 75 by the formulas 
of this article. 

Ans j — 430 pounds per square inch. 

(compression). 


70 “ 


a a 


u 


STRENGTH OF COLUMNS. 

A stick of timber, a bar of iron, etc., when used to sustain 
end loads which act lengthwise of the pieces, are called columns, 
posts, or struts if they are so long that they would bend before 
breaking. When they are so short that they would not bend 
before breaking, they are called short blocks, and their compres¬ 
sive strengths are computed by means of equation 1. The strengths 
of columns cannot, however, be so simply determined, and we now 
proceed to explain the method of computing them. 

77. End Conditions. The strength of a column depends in 
part on the way in which its ends bear, or are joined to other 
parts of a structure, that is, on its “ end conditions.” There are 
practically but three kinds of end conditions, namely: 

1. “Hinge” or “pin” ends, 

2. “ Flat ” or “ square ” ends, and 

3. “Fixed” ends. 

(1) When a column is fastened to its support at one end by 
means of a pin about which the column could rotate if the other 
end were free, it is said to be “ hinged” or “ pinned” at the 
former end. Bridge posts or columns are often hinged at the ends. 

(2) A column either end of which is flat and perpendicular 
to its axis and bears on other parts of the structure at that surface, 
is said to be “ flat ” or “ square” at that end. 














86 


STRENGTH OF MATERIALS 


(3) Columns are sometimes riveted near their ends directly 
to other parts of the structure and do not bear directly on their 
ends; such are called “fixed ended.” A column which bears on its 
flat ends is often fastened near the ends to other parts of the struc¬ 
ture, and such an end is also said to be “ fixed.” The fixing of an 
end of a column stiffens and therefore strengthens it more or less, 
but the strength of a column with fixed ends is computed as 
though its ends were fiat. Accordingly we have, so far as strength 

is concerned, the following classes of columns : 

% 

78. Classes of Columns. ( 1 ) Both ends hinged or pinned; 

(2) one end hinged and one flat; (3) both ends flat. 

Other things being the same, columns of these three classes 
are unequal in strength. Columns of the first class are the 
weakest, and those of the third class are the strongest. 


IB 

B 

A 



A A 



A 








l B 

B 


6- b 

Fig. 46. 

79. Cross=sections of Columns. Wooden columns are usu¬ 
ally solid, square, rectangular, or round in section; but sometimes 
they are “ built up ” hollow. Cast-iron columns are practically 
always made hollow, and rectangular or round in section. Steel 
columns are made of single rolled shapes—angles, zees, channels, 
etc.; but the larger ones are usually “ built up ” of several shapes. 
Fig. 46, <2, for example, represents a cross-section of a “ Z-bar” 
column; and Fig. 46, b, that of a “channel” column. 

80 . Radius of Gyration. There is a quantity appearing in 
almost all formulas for the strength of columns, which is called 
“ radius of gyration.” It depends on the form and extent of the 
cross-section of the column, and may be defined as follows: 
























STRENGTH OF MATERIALS 


87 


The radius of gyration of any plane figure (as the section of a column) 
with respect to any line, is such a length that the square of this length mul¬ 
tiplied by the area of the figure equals the moment of inertia of the figure 
with respect to the given line. 

Thus, if A denotes the area of a figure; I, its moment of in¬ 
ertia with respect to some line; and r, the radius of gyration 
with respect to that line; then 

r 2 A = I; or r = l/l A. (9) 

In the column formulas, the radius of gyration always refers to an 
axis through the center of gravity of the cross-section, and usually 
to that axis with respect to which the radius of gyration (and mo¬ 
ment of inertia) is least. (For an exception, see example 3, 
Art. 83.) Hence the radius of gyration in this connection is often 
called for brevity the “ least radius of gyration,” or simply the 
“ least radius.” 

Examples. 1. Show that the value of the radius of gyration 
given for the square in Table A, page 52, is correct. 

The moment of inertia of the square with respect to the axis 
is iSince A = a 2 , then, by formula 9 above, 



2. Prove that the value of the radius of gyration given for 
the hollow square in Table A, page 54, is correct. 

The value of the moment of inertia of the square with respect 
to the axis is ^ (a 4, - a 4 '). Since A = dr - a 2 , 



EXAflPLE FOR ^PRACTICE. 

Prove that the values of the radii of gyration of the other fig¬ 
ures given in Table A, page 52, are correct. The axis in each 
case is indicated by the line through the center of gravity. 

8 i. Radius of Gyration of Built=up Sections. The radius of 
gyration of a built-up section is computed similarly to that of any 
other figure. First, we have to compute the moment of inertia of 
















88 


STRENGTH OF MATERIALS 


the section, as explained in Art. 54; and then we use formula 9, as 
in the examples of the preceding article. 

Example. It is required to compute the radius of gyration 
of the section represented in Fig. 30 (page 52) with respect to the 
axis AA. 

In example 1, Art. 54, it is shown that the moment of inertia 
of the section with respect to the axis AA is 429 inches 4 . The 
area of the whole section is 

2 X 6.03 + 7 = 19.06; 
hence the radius of gyration ris 

r — — = 4.74 inches. 

M 19.06 


EXAMPLE FOR PRACTICE. 

Compute the radii of gyration of the section represented in 
Fig. 31, a , with respect to the axes AA and BB. (See examples 
for practice 1 and 2, Art. 54.) 

. ( 2.87 inches. 

Ans ‘ 12.09 “ 

82. Kinds of Column Loads. When the loads applied to a 
column are such that their resultant acts through the center of 
gravity of the top section and along the axis of the column, the 
column is said to be centrally loaded. When the resultant of the 
loads does not act through the center of gravity of the top 
section, the column is said to be eccentrically loaded. All the 
following formulas refer to columns centrally loaded. 

83. Rankine’s Column Formula. When a perfectly straight 
column is centrally loaded, then, if the column does not bend and 
if it is homogeneous, the stress on every cross-section is a uniform 
compression. If P denotes the load and A the area of the cross- 
section, the value of the unit-compression is P -r- A. 

On account of lack of straightness or lack of uniformity in 
material, or failure to secure exact central application of the load, 
the load P has what is known as an “ arm ” or “ leverage ” and 
bends the column more or less. There is therefore in such a 
column a bending or flexural stress in addition to the direct com¬ 
pressive stress above mentioned; this bending stress is compressive 





STRENGTH OF MATERIALS 


89 


on the concave side and tensile on the convex. The value of the 
stress per unit-area (unit-stress) on the fibre at the concave side, 
according to equation 6, is I, where M denotes the bending 

moment at the section (due to the load on the column), c the 
distance from the neutral axis to the concave side, and I the 
moment of inertia of the cross-section with respect to the neutral 
axis. (Notice that this axis is perpendicular to the plane in 
which the column bends.) 

The upper set of arrows (Fig. 47) represents the direct com¬ 
pressive stress; and the second set the bending stress if the load 
is not excessive, so that the stresses are within the elastic limit of 
the material. The third set represents the combined stress that 
actually exists on the cross-section. The greatest combined unit- 
stress evidently occurs on the fibre at the concave side and where 

the deflection of the column is greatest. The 
stress is compressive, and its value S per unit- 
area is given by the formula, 

M o 





ffTTT^ ' J 




mm: 


IPft 


Fig. 47. 


S = A+ x 

Now, the bending moment at the place of 
greatest deflection equals the product of the 
load P and its arm (that is, the deflection). 
Calling the deflection d , we have M = P d\ and 
this value of M, substituted in the last equa 
tion, gives 

c P P do 

s= x+ 


Let r denote the radius of gyration of the cross-section with respec 
to the neutral axis. Then I = Ar* (see equation 9); and this 
value, substituted in the last equation, gives 


S = x + 


P do P . dc 


A r 2 A 


According to the theory of the stiffness of beams on end sup¬ 
ports, deflections vary directly as the square of the length Z, and in¬ 
versely as the distance c from the neutral axis to the remotest fibie 
of the cross-section. Assuming that the deflections of columns 



































90 


STRENGTH OF MATERIALS 


follow the same laws, we may write d = Jc (l 2 -f- c), where & is 
some constant depending on the material of the column and on the 
end conditions. Substituting this value for d in the last equation, 
we find that 


and 


S : 

P _ 
X ~~ 

p = 


p l 2 ' 

x ( 1 + 
s 

l 2 ’ 

1 + ^ r l 


SA 


l 2 

1 + 1c X 





Each of these (usually the last) is known as “ Rankme’s formula.” 

For mild-steel columns a certain large steel company uses S = 50,000 
pounds per square inch, and the following values of k: 

1. Columns with two pin ends, k = 1 -*- 18,000. 

2. “ “ one flat and one pin end, k = 1 -s- 24,000. 

3. “ “ both ends flat, k = 1 -4- 36,000. 

With these values of S and k , P of the formula means the ultimate load, 
that is, the load causing failure. The safe load equals P divided by the 
selected factor of safety—a factor of 4 for steady loads, and 5 for moving 
loads, being recommended by the company referred to. The same unit is to 
be used for l and r. 


Cast-iron columns are practically always made hollow with 
comparatively thin walls, and are usually circular or rectangular 
in cross-section. Tbs following modifications of Rankine’s formula 
are sometimes used: 


For circular sections, 


For rectangular sections, 


p 

80,000 

A 1- 

i p 


r 800 d 2 

p 

80,000 

A ' 1. 

+ 11 


1,000 d 2 



In these formulas d denotes the outside diameter of the circular sec¬ 
tions or the length of the lesser side of the rectangular sections. The same 
unit is to be used for Z and d. 


Examples. 1. A 40-pound 10-inch steel I-beam 8 feet 
long is used as a flat-ended column. Its load being 100,000 
pounds, what is its factor of safety ? 

Obviously the column tends to bend in a plane perpendicular 
to its web. Hence the radius of gyration to be used is the one 











STRENGTH OF MATERIALS 


91 


with respect to that central axis of the cross-section which is in 
the web, that is, axis 2 - 2 (see figure accompanying table, page 72). 
The moment of inertia of the section with respect to that axis, 
according to the table, is 9.50 inches 4 ; and since the area of the 
section is 11.76 square inches, 

= 1L76 = °' 80S ' 


Now, 1 = 8 feet = 96 inches; and since Jc = 1 -f- 36,000, and S = 
50,000, the breaking load for this column, according to Rankine’s 
formula, is 


50,000 X 11.76 
~ 96 * 

1 + 36,000 X 0.808. 


446,790 pounds. 


Since the factor of safety equals the ratio of the breaking load to 
the actual load on the column, the factor of safety in this case is 


446,790 

T00.000 


4.5 (approx.). 


2. What is the safe load for a cast-iron column 10 feet long 
with square ends and a hollow rectangular section, the outside 
dimensions being 5x8 inches; the inner, 4x7 inches; and the 
factor of safety, 6 ? 

In this case Z — 10 feet = 120 inches; A = 5 X 8-4 X 7 
= 12 square inches; and d — 5 inches. Hence, according to 
formula 10', for rectangular sections, the breaking load is 

80,000 X 12 


1 + 


120 2 


610,000 pounds. 


1,000 X 5 2 

Since the safe load equals the breaking load divided by the factor 
of safety, in this case the safe load equals 

610,000 


6 


101,700 pounds. 


3. A channel column (see Fig. 46, b) is pin-ended, the pins 
being perpendicular to the webs of the channel (represented by 
AA in the figure), and its length is 16 feet (distance between axes 














92 


STRENGTH OF MATERIALS 


of the pins). If the sectional area is 23.5 square inches, and the 
moment of inertia with respect to AA is 386 inches 4 and with 
respect to BB 214 inches 4 , what is the safe load with a factor of 
safety of 4 ? 

The column is liable to bend in one of two ways, namely, in 
the plane perpendicular to the axes of the two pins, or in the plane 
containing those axes. 

^1) For bending in the first plane, the strength of the col¬ 
umn is to be computed from the formula for a pin-ended column. 
Hence, for this case, r 2 = 386 - 5 - 23.5 = 16; and the breaking 
load is 


P = 


50,000 X 23.5 


1 + 


(16 X 12) 2 


1,041,600 pounds. 


18,000 X 16 

The safe load for this case equals — ^ = 260,400 pounds. 


(2) If the supports of the pins are rigid, then the pins 
stiffen the column as to bending in the plane of their axes, and the 
strength of the column for bending in that plane should be com¬ 
puted from the formula for the strength of columns with flat ends. 
Hence, r 2 = 214 -s- 23.5 = 9.11, and thebreaking load is 


50,000 X 23.5 


1 + o77 


(16 X 12) 2 


= 1,056,000 pounds. 


36,000 X 9.11 


The safe load for this case equals - = 264,000 pounds. 


EXAMPLES FOR PRACTICE. 

1. A 40-pound 12-inch steel I-beam 10 feet long is used as 
a column with flat ends sustaining a load of 100,000 pounds. 
What is its factor of safety? 

Ans. 4.1 

2. A cast-iron column 15 feet long sustains a load of 
150,000 pounds. Its section being a hollow circle, 9 inches out¬ 
side and 7 inches inside diameter, what is the factor of safety? 

Ans. 8.9 

3. A steel Z-bar column (see Fig. 46, a) is 24 feet long and 
has square ends; the least radius of gyration of its cross-section is 











STRENGTH OF MATERIALS 


93 


3.1 inches; and the area of the cross-section is 24.5 square inches. 
What is the safe load for the column with a factor of safety of 4? 

Ans. 247,000 pounds. 

4. A cast-iron column 13 feet loner has a hollow circular 

O 

cross-section 7 inches outside and 54 inches inside diameter. 
What is its safe load with a factor of safety of 6? 

Ans. 121,142 pounds. 

5. Compute the safe load for a 40-pound 12-inch steel 
I-beam used as a column with flat ends, its length being 17 feet. 
Use a factor of safety of 5. 

Ans. 52,470 pounds. 
84. Graphical Representation of Column Formulas. Col¬ 
umn (and most other engineering) formulas can be represented 
graphically. To represent Rankine’s formula for flat-ended mild- 
steel columns, 

P 50,000 

A (l -s- r ) 2 ’ 

1 4 - -- — 

1 + 36,000 

we first substitute different values of l -t- r in the formula, and 
solve for P -h A. Thus we find, when 

l + r = 40, P + A = 47,900 ; 
l'+r = 80, P + A = 42,500; 
l + r = 120, P + A = 35,750 ; 
etc., etc. 

Now, if these values of l h- r be laid off by some scale on a line 
from O, Fig. 48, and the corresponding values of P A be laid 

P-A 



off vertically from the points on the line, we get a series of points 
as a, b, c, etc.; and a smooth curve through the points a, b , c t 










94 


STRENGTH OF MATERIALS 


etc., represents the formula. Such a curve, besides representing 
the formula to one’s eye, can be used for finding the value of 
P - 7 - A for any value of l -s- r; or the value of / -r- r for any value 
of P -s- A. The use herein made is in explaining other column 
formulas in succeeding articles. 

85 . Combination Column Formulas. Many columns have 
been tested to destruction in order to discover in a practical way 
the laws relating to the strength of columns of different kinds. 
The results of such tests can be most satisfactorily represented 
graphically by plotting a point in a diagram for each test. Thus, 
suppose that a column whose l - 5 - r was 80 failed under a load of 
276,000 pounds, and that the area of its cross-section was 7.12 
square inches. This test would be represented by laying off O a, 
Fig. 49, equal to 80, according to some scale; and then ab equal to 
276,000 H- 7.12 (P -r- A), according to some other convenient 
scale. The point b would then represent the result of this par¬ 
ticular test. All the dots in the figure represent the way in which 
the results of a series of tests appear when plotted. 

It will be observed at once that the dots do not fall upon any 
one curve, as the curve of Rankine’s formula. Straight lines and 


P4-A 

50000- 
40000- 
30000 - 
20000 - 

10000 





_ 

100 

Fig. 49. 




200 


1 1* r 

300 


curves simpler than the curve of Rankine’s formula have been 
fitted to represent the average positions of the dots as determined 
by actual tests, and the formulas corresponding to such lines have 
been deduced as column formulas. These are explained in the 
following articles. 

O 


86. Straight-Line and Euler Formulas. It occurred to Mr. 
T. H. Johnson that most of the dots corresponding to ordinary 







STRENGTH OF MATERIALS 


95 


lengths of columns agree with a straight line just as well as with 
a curve. He therefore, in 1886, made a number of such plats or 
diagrams as Fig. 49, fitted straight lines to them, and deduced the 
formula corresponding to each line. These have become known 
as “ straight-line formulas,” and their general form is as follows: 

_=S-m—, (II) 

P, A, l, and shaving meanings as in Rankine’s formula (Art. 83), 
and S and m being constants whose values according to Johnson 
are given in Table E below. 

For the slender columns, another formula (Euler’s, long since 
deduced) was used by Johnson. Its general form is— 

P n 

X = {I ~ rf 

n being a constant whose values, according to Johnson, are given 
in the following table: 

TABLE E. 

Data for Mild-Steel Columns. 



S 

m 

Limit ( l -i- r) 

n 

Hinged ends. 

52,500 

220 

160 

444,000,000 

Flat ends. 

52,500 

180 

195 

666,000,000 


The numbers in the fourth column of the table mark the point of divi¬ 
sion between columns of ordinary length and slender columns. For the 
former kind, the straight-line formula applies; and for the second, Euler’s. 
That is, if the ratio l r for a steel column with hinged end, for example, is 
less than 160, we must use the straight-line formula to compute its safe load, 
factor of safety, etc.; but if the ratio is greater than 160, we must use Euler’s 
formula. 

For cast-iron columns with flat ends, S = 31,000, and m = 88; and since 
they should never be used “slender,” there is no use of Euler’s formula for 
cast-iron columns. 

The line AB, Fig. 50, represents Johnson’s straight-line for¬ 
mula; and BC, Euler’s formula. It will be noticed that the two 
lines are tangent; the point of tangency corresponds to the “lim¬ 
iting value ” l -7- r y as indicated in the table. 

« 

Examples. 1. A 40-pound 10-inch steel I-beam column 8 



















96 


STRENGTH OF MATERIALS 


feet long sustains a load of 100,000 pounds, and the ends are flat. 
Compute its factor of safety according to the methods of this 
article. 

The first thing to do is to compute the ratio l -r- r for the 
column, to ascertain whether the straight-line formula or Euler’s 



formula should be used. From Table C, on page 70, we find that 
the moment of inertia of the column about the neutral axis of 
its cross-section is 9.50 inches 4 , and the area of the section is 
11.76 square inches. Hence 

9.50 

r 2 = ' - j - A - = 0.81: or r — 0.9 inch. 

11 . 


Since l — 8 feet = 96 inches, 

J _ 96 _ 
r 0.9 



This value of l -s- r is less than the limiting value (195) indicated 
by the table for steel columns with flat ends (Table E, p. 97), and 
we should therefore use the straight-line formula; hence 

i-fb? = 52,500 - 180 X 106-?-5 

11.7o o 

or, P = 1J.76 (52,500 - 180 X 106-?-) = 391,600 pounds. 

O 

This is the breaking load for the column according to the straight* 
line formula; hence the factor of safety is 

391,600 _ 

100,000 - 









STRENGTH OF MATERIALS 


97 


2. Suppose that the length of the column described in the 
preceding example were 16 feet. MTiat would its factor of safety be? 

Since l = 16 feet = 192 inches; and, as before, r = 0.9 
inch, l r = 213J. This value is greater than the limiting 

value (195) indicated by Table E (p. 97) for flat-ended steel col¬ 
umns; hence Euler’s formula is to be used. Thus 


P _ 666,000,000 

11.76 “ (213^) 2 ; 


or 


P = 


11.76 X 666,000,000 


(213 iY 

This is the breaking load; hence the factor of safety is 


= 172,100 pounds. 


172,100 

100,000 * 


3. "What is the safe load for a cast-iron column 10 feet long 
with square ends and hollow rectangular section, the outside 
dimensions being 5x8 inches and the inside 4x7 inches, with a 
factor of safety of 6 ? 

Substituting in the formula for the radius of gyration given 
in Table A, page 52, we get 





8 X 5 3 - 7 X 4 3 
12 (8 X 5 - 7 X 4) 


= 1.96 inches. 


Since l = 10 feet = 120 inches, 


l 


r 


120 

L96 


61.22 


According to the straight-line formula for cast iron, A being 
equal to 12 square inches, 

T = 34,000 - 88 X 61.22; 

or, P = 12 (34,000 - 88 X 61.22) = 343,360 pounds. 

This being the breaking load, the safe load is 

343,360 now j 

-1- = 57,227 pounds. 

6 










98 


STRENGTH OF MATERIALS 


EXAMPLES FOR PRACTICE. 

1. A 40-pound 12-inch steel I-beam 10 feet long is used as 
a flat-ended column. Its load being 100,000 pounds, compute 
the factor of safety by the formulas of this article. 

Ans. 3.5 

2. A cast-iron column 15 feet long sustains a load of 
150,000 pounds. Its section being a hollow circle of 9 inches 
outside and 7 inches inside diameter, compute the factor of safety 
by the straight-line formula. 

J Ans. 4.8 

3. A steel Z-bar column (see Fig. 46, a) is 24 feet long 
and has square ends; the least radius of gyration of its cross- 
section is 3.1 inches; and the area of the cross-section is 24.5 
square inches. Compute the safe load for the column by the 
formulas of this article, using a factor of safety of 4. 

Ans. 219,000 pounds. 

4. A hollow cast-iron column 13 feet long has a circular 
cross-section, and is 7 inches outside and 54 inches inside in 
diameter. Compute its safe load by the formulas of this article, 
using a factor of safety of 6. 

Ans. 68,500 pounds 

5. Compute by the methods of this article the safe load for 
a 40-pound 12-inch steel I-beam used as a column with flat ends, 
if the length is 17 feet and the factor of safety 5. 

Ans. 35,100 pounds. 

87 . ParaboIa=EuIer Formulas. As better fitting the results 
of tests of the strength of columns of “ ordinary lengths,” Prof. 
J. B. Johnson proposed (1892) to use parabolas instead of straight 
lines. The general form of the “ parabola formula ” is 

= (13) 

P, A, l and r having the same meanings as in Rankine’s formula, 
Art. 83; and S and m denoting constants whose values, according 
to Professor Johnson, are given in Table F below. 

Like the straight-line formula, the parabola formula should 
not be used for slender columns, but the following (Euler’s) is 
applicable: 


STRENGTH OF MATERIALS 


99 


i n 

X = (Th- t) 2 ' 


the values of n (Johnson) being given in the following table: 

TABLE F. 

Data for flild Steel Columns. 



S 

m 

Limit (Z -4- r) 

n 

Hinged ends....... 

42,000 

0.97 

150 

456,000,000 

Flat ends. 

42,000 

0.62 

190 

712,000,000 


The point of division between columns of ordinary length and slender 
columns is given in the fourth column of the table. That is, if the ratio Z-j-r 
for a column with hinged ends, for example, is less than 150, the parabola 
formula should be used to compute the safe load, factor of safety, etc.; but 
if the ratio is greater than 150, then Euler’s formula should be used. 

The line AB, Fig. 51, represents the parabola formula; and the line 
BC, Euler’s formula. The two lines are tangent, and the point of tangency 
corresponds to the 1 ‘ limiting value ’ ’ Z-Pr of the table. 

For wooden columns square in cross-section, it is convenient to replace 
r by d, the latter denoting the length of the sides of the square. The formula 
becomes 


— = S - m 

A 



S and m for flat-ended columns of various kinds of wood having the follow¬ 
ing values according to Professor Johnson: 

For White pine, S=2,500, m = 0.6; 

“ Short-leaf yellow pine, S=3,300, m = 0.7; 

‘ ‘ Long-leaf yellow pine, S=4,000, w = 0.8; 

‘ 1 White oak, S=3,500, m — 0.8. 

The preceding formula applies to any wooden column whose ratio, Z-Pd, 
is less than 60, within which limit columns of practice are included. 



) 

* 


* ' > 




















100 


STRENGTH OF MATERIALS 


8 feet long sustains a load of 100,000 pounds, and its ends are flat. 
Compute its factor of safety according to the methods of this 
article. 

The first thing to do is to compute the ratio l -4- r for the 
column, to ascertain whether the parabola formula or Euler’s for¬ 
mula should be used. As shown in example 1 of the preceding 
article, l ~ r = 106§. This ratio being less than the limiting 
value, 190, of the table, we should use the parabola formula. 
Hence, since the area of the cross-section is 11.76 square inches 
(see Table C, page 70), 

—— = 42,000 - 0.62 (106SV: 

11.76 ’ v SJ ’ 


or, P = 11.76 [42,000 - 0.62 (106§) 2 ] = 410,970 pounds. 

This is the breaking load according to the parabola formula; hence 
the factor of safety is 

410,970 _ ± i 

100,000 


2. A white pine column 10 X 10 inches in cross-section and 
18 feet long sustains a load of 40,000 pounds. What is its factor 
of safety ? 

The length is 18 feet or 216 inches; hence the ratio l -f- d = 
21.6, and the parabola formula is to be applied. 

Now, since A = 10 X 10 = 100 square inches, 

P 

jqq = 2,500 - 0.6 X 21.6*; 

or, P = 100 (2,500 - 0.6 X 21.6 2 ) = 222,000 pounds. 

This being the breaking load according to the parabola formula, 
the factor of safety is 

222,000 

40,000 = 5 ' 5 

3. What is the safe load for a long-leaf yellow pine column 
12 X 12 inches square and 30 feet long, the factor of safety 
being 5 ? 

The length being 30 feet or 360 inches, 


l 

d 








STRENGTH OF MATERIALS 


101 


hence the parabola formula should be used. Since A = 12 X 12 
= 144 square inches, 

P 

— = 4,000 - 0.8 X 30 2 ; 

or, P = 144 (4,000 - 0.8 X 30 2 ) = 472,320 pounds. 

This being the breaking load according to the parabola formula, 
the safe load is 


472,320 

5 


= 94,465 pounds. 


EXAMPLES FOR PRACTICE. 

1. A 40-pound 12-inch steel I-beam 10 feet long is used as 
a flat-ended column. Its load being 100,000 pounds, compute its 
factor of safety by the formulas of this article. 

Ans. 3.8 

2. A white oak column 15 feet long sustains a load of 
30,000 pounds. Its section being 8x8 inches, compute the 
factor of safety by the parabola formula. 

Ans. 6.6 

3. A steel Z-bar column (see Fig. 46, a') is 24 feet long and 
has square ends; the least radius of gyration of its cross-section 
is 3.1 inches; and the area of its cross-section is 24.5 square 
inches. Compute the safe load for the column by the formulas 
of this article, using a factor of safety of 4. 

Ans. 224,500 pounds. 

4. A short-leaf yellow pine column 14 X 14 inches in sec¬ 
tion is 20 feet long. What load can it sustain, with a factor of 
safety of 6 ? 

Ans. 101,100 pounds. 

88. “ Broken Straight=Line ” Formula. A large steel com¬ 

pany computes the strength of its flat-ended steel columns by two 
formulas represented by two straight lines AB and BC, Fig. 52. 
The formulas are 

X = 48,000, 

P l 

and -j = 68,400 - 228 -, 

P, A, l, and r having the same meanings as in Art. 88. 




102 


STRENGTH OF MATERIALS 


The point B corresponds very nearly to the ratio l -r- r = 90. 
Hence, for columns for which the ratio l -s- r is less than 90, the 
first formula applies; and for columns for which the ratio is 
greater than 90, the second one applies. The point C corre¬ 
sponds to the ratio l -f- r = 200, and the second formula does not 
apply to a column for which l . -s- r is greater than that limit. 



The ratio l -s- r for steel columns of practice rarely exceeds 150, 
and is usually less than 100. 

Fig. 53 is a combination of Figs. 49, 50, 51 and 52, and 
represents graphically a comparison of the Rankine, straight-line, 
Euler, parabola-Euler, and broken straight-line formulas for flat- 
ended mild-steel columns, It well illustrates the fact that our 
knowledge of the strength of columns is not so exact as that, for 
example, of the strength of beams. 



89 . Design of Columns. All the preceding examples relat¬ 
ing to columns were on either ( 1 ) computing the factor of safety 









STRENGTH OF MATERIALS 


103 


of a given loaded column, or (2) computing the safe load for a 
given column. A more important problem is to design a column 
to sustain a given load under given conditions. A complete dis¬ 
cussion of this problem is given in a later paper on design. We 
show here merely how to compute the dimensions of the cross- 
section of the column after the form of the cross-section has been 
decided upon. 

In only a few cases can the dimensions be computed directly 
(see example 1 following), but usually, when a column formula is 
applied to a certain case, there will be two unknown quantities in 
it, A and r or d. Such cases can best be solved by trial (see 
examples 2 and 3 below). 

Example. 1. What is the proper size of white pine column 
to sustain a load of 80,000 pounds with a factor of safety of 5, 
when the length of the column is 22 feet ? 

We use the parabola formula (equation 13). Since the safe 
load is 80,000 pounds and the factor of safety is 5, the breaking 
load P is 

80,000 X 5 = 400,000 pounds. 

The unknown side of the (square) cross-section being denoted by 
d , the area A is d 2 . Hence, substituting in the formula, since l 
= 22 feet == 264 inches, we have 

400 ’ 000 = 2,500 - 0.6 2 ^. 

<P d 2 

Multiplying both sides by d 2 gives 

400,000 = 2,500 d 2 - 0.6 X 264 2 , 
or 2,500 d 2 = 400,000 + 0.6 X 264 2 = 441,817.6. 

Hence d 2 = 176.73, or d = 13.3 inches. 

2. What size of cast-iron column is needed to sustain a load 
of 100,000 pounds with a factor of safety of 10, the length of the 
column being 14 feet ? 

We shall suppose that it has been decided to make the cross- 
section circular, and shall compute by Rankine’s formula modified 
for cast-iron columns (equation 10'). The breaking load for the 
column would be 










104 


STRENGTH OF MATERIALS 


100,000 X 10 = 1,000,000 pounds. 

The length is 14 feet or 168 inches; hence the formula Decomes 

« 

1,000,000 80,000 
A " 168* ; 

1 + 800d 2 


or, reducing by dividing both sides of the equation by 10,000, and 
then clearing of fractions, we have 

100 0 + 800fld = 8A ‘ 


There are two unknown quantities in this equation, d and A, and 
we cannot solve directly for them. Probably the best way to pro¬ 
ceed is to assume or guess at a practical value of d , then solve for 
A, and finally compute the thickness or inner diameter. Thus, let 
us try d equal to 7 inches, first solving the equation for A as far 
as possible. Dividing both sides by 8 we have 

168 2 -I 

800^ 2 J’ 



and, combining, 


A = 12.5 + 


441 

d 2 ' 


Now, substituting 7 for d , we have 

441 


A = 12.5 + -jg- = 21.5 square inches. 


The area of a hollow circle whose outer and inner diameters are 
d and d x respectively, is 0.7854 ( d 2 - d 2 ). Hence, to find the inner 
diameter of the column, we substitute 7 for d in the last expres¬ 
sion, equate it to the value of A just found, and solve for d . Thus, 

0.7854 (49 - d 2 ) = 21.5* 

hence 


49 - d 2 = 


21.5 

077854 “ 


27.37; 


and d* = 49 - 27.37 = 21.63 or d x = 4.65. 

This value of d makes the thickness equal to 

J (7 - 4.65) = 1.175 inches, 


f 











STRENGTH OF MATERIALS 


105 


which is safe. It might be advisable in an actual case to try 
d equal to 8 repeating the computation.* 

EXAMPLE FOR PRACTICE. 

1. What size of white oak column is needed to sustain a load 
of 45,000 pounds with a factor of safety of 6, the length of the 
column being 12 feet. 

Ans. d = 84, practically a 10 X 10-inch section 

STRENGTH OF SHAFTS. 

A shaft is a part of a machine or system of machines, and is 
used to transmit power by virtue of its torsional strength, or resist¬ 
ance to twisting. Shafts are almost always made of metal and are 
usually circular in cross-section, being sometimes made hollow. 

90 . Twisting Moment. Let AF, Fig. 54, represent a shaft 
with four pulleys 011 it. Suppose that D is the driving pulley 
and that B, C and E are pulleys from which power is taken off to 
drive machines. The portions of the shafts between the pulleys 



are twisted when it is transmitting power; and by the twisting 
moment at any cross-section of the shaft is meant the algebraic 
sum of the moments of all the forces acting on the shaft on either 


*Note. The structural steel handbooks contain extensive tables by- 
means of which the design of columns of steel or cast iron is much facilitated. 
The difficulties encountered in the use of formulae are well illustrated in this 
example. 




106 


STRENGTH OF MATERIALS 


side of the section, the moments being taken with respect to the 
axis of the shaft. Thus, if the forces acting on the shaft (at the 
pulleys) are P 4 , P 2 , P 3 , and P 4 as shown, and if the arms of the 
forces or radii of the pulleys are a v a 2 , a 3 , and a i respectively, then 
the twisting moment at any section in 

BC is Pj a l9 

CD is Pj a x + P 2 a 2 , 

DE is Pj a x -j- P 2 ^2 — P 3 

Like bending moments, twisting moments are usually ex¬ 
pressed in inch-pounds. 

Example. Let a x = a 2 = a x = 15 inches, a 3 = 30 inches, 
Pj = 400 pounds, P 2 = 500 pounds, P 3 = 750 pounds, and P 4 = 
600 pounds.* What is the value of the greatest twisting moment 
in the shaft ? 

At any section between the first and second pulleys, the 
twisting moment is 

400 X 15 = 6,000 inch-pounds; 

at any section between the second and third it is 

400 X 15 + 500 X 15 = 13,500 inch-pounds; and 
at any section between the third and fourth it is 
400 X 15 + 500 X 15 - 750 X 30 = - 9,000 inch-pounds. 
Hence the greatest value is 13,500 inch-pounds. 

91. Torsional Stress. The stresses in a twisted shaft are 
called “ torsional” stresses. The torsional stress on a cross-section 
of a shaft is a shearing stress, as in the case illustrated by Fig. 55, 
which represents a flange coupling in a shaft. Were it not for 
the bolts, one flange would slip over the other when either part 
of the shaft is turned; but the bolts prevent the slipping. Obvi¬ 
ously there is a tendency to shear the bolts off unless they are 
screwed up very tight; that is, the material of the bolts is sub¬ 
jected to shearing stress. 

Just so, at any section of the solid shaft there is a tendency 
for one part to slip past the other, and to prevent the slipping or 

* Note. These numbers were so chosen that the moment of P (driving 
moment) equals the sum of the moments of the other forces. This is always 
the case in a shaft rotating at constant speed; that is, the power given the 
shaft equals the power taken off. 















STRENGTH OF MATERIALS 


107 


shearing of the shaft, there arise shearing stresses at all parts of 
the cross-section. The shearing stress on the cross-section of a 
shaft is not a uniform stress, its value per unit-area being zero at 
the center of the section, and increasing toward the circumference. 
In circular sections, solid or hollow, the shearing stress per unit- 
area (unit-stress) varies directly as the distance from the center 
of the section, provided the elastic limit is not exceeded. Thus, 
if the shearing unit-stress at the circumference of a section is 




Fig. 55. 


1,000 pounds per square inch, and the diameter of the shaft is 
2 inches, then, at inch from the center, the unit-stress is 500 
pounds per square inch; and at | inch from the center it is 250 
pounds per square inch. In Fig. 55 the arrows indicate the 
values and the directions of the shearing stresses on very small 
portions of the cross-section of a shaft there represented. 

92. Resisting Moment. By “resisting moment” at a sec¬ 
tion of a shaft is meant the sum of the moments of the shearing 
stresses on the cross-section about the axis of the shaft. 

Let S s denote the value of the shearing stress per unit-area 
(unit-stress) at the outer points of a section of a shaft; d the 
diameter of the section (outside diameter if the shaft is hollow); 
and d x the inside diameter. Then it can be shown that the re¬ 
sisting moment is: 

For a solid section, 0.1963 S s d z \ 

T , , „ .. 0.1963 S (d* - (I*), 

lor a hollow section, --^- 

93. Formula for the Strength of a Shaft. As in the case 

















108 


STRENGTH OP MATERIALS 


* 


of beams, the resisting moment equals the twisting moment at 
any section. If T be used to denote twisting moment, then we 
have the formulas : 


For solid circular shafts, 0.1963 S s d z = T; ) 

For hollow circular shafts, °- 19C3 ^ (^ - "V) = T- j- (15) 


In any portion of a shaft of constant diameter, the unit- 
shearing stress S s is greatest where the twisting moment is greatest. 
Hence, to compute the greatest unit-shearing stress in a shaft, 
we first determine the value of the greatest twisting moment, 
substitute its value in the first or second equation above, as the 
case may be, and solve for S s . It is customary to express T in 
inch-pounds and the diameter in inches, S s then being in pounds 
per square inch. 

Examples. 1. Compute the value of the greatest shearing 
unit-stress in the portion of the shaft between the first and second 
pulleys represented in Fig. 54, assuming values of the forces and 
pulley radii as given in the example of article 90. Suppose also 
that the shaft is solid, its diameter being 2 inches. 

The twisting moment T at any section of the portion between 
the first and second pulleys is 6,000 inch-pounds, as shown in the 
example referred to. Hence, substituting in the first of the two 
formulas 15 above, we have 


or. 


S s = 


0.1963 S s X 2 3 = 6,000; 

6,000 

(ri903 ' x '~ 8 = pounds per square inch. 


This is the value of the unit-stress at the outside portions of all 
sections between the first and second pulleys. 

2. A hollow shaft is circular in cross-section, and its outer 
and inner diameters are 16 and 8 inches respectively. If the 
working strength of the material in shear is 10,000 pounds per 
square inch, what twisting moment can the shaft safely sustain ? 

The problem requires that we merely substitute the values of 
S a , <7, and d A in the second of the above formulas 15, and solve for 
T. Thus, 


0.1963 x 10,000 (16* - 8 4 ) 

16 


= 7,537,920 inch-pounds. 


T 





STRENGTH OF MATERIALS 


109 


EXAMPLES FOR PRACTICE. 


1. Compute the greatest value of the shearing unit-stress in 
the shaft represented in Fig. 54, using the values of the forces 
and pulley radii given in the example of article 90, the diameter 
of the shaft being 2 inches. 

Ans. 8,595 pounds per square inch 

2. A solid shaft is circular in cross-section and is 9.G inches 
in diameter. If the working strength of the material in shear is 
10,000 pounds per square inch, how large a twisting moment can 
the shaft safely sustain? (The area of the cross-section is practically 
the same as that of the hollow shaft of example 2 preceding.) 

Ans. 1,736,736 inch-pounds. 

94. Formula for the Power Which a Shaft Can Transmit. 
The power that a shaft can safely transmit depends on the shear¬ 
ing working strength of the material of the shaft, on the size of 
the cross-section, and on the speed at which the shaft rotates. 

Let H denote the amount of horse-power; S s the shearing 
working strength in pounds per square inch; d the diameter 
(outside diameter if the shaft is hollow) in inches; d x the inside 
diameter in inches if the shaft is hollow; and n the number of 
revolutions of the shaft per minute. Then the relation between 
power transmitted, unit-stress, etc., is: 


For solid shafts, H = 
For hollow shafts, H — 


S s rP n } 

321,000 ’ 

S s (d‘ - d‘) n 
321,000 d 



Examples. 1. What horse-power can a hollow shaft 16 
inches and 8 inches in diameter safely transmit at 50 revolutions 
per minute, if the shearing working strength of the material is 
10,000 pounds per square inch? 

We have merely to substitute in the second of the two for¬ 
mulas 16 above, and reduce. Thus, 



10,000 (16* - 8 4 ) 50 
321,000 X 16 “ 


6,000 horse-power (nearly). 


2. What size of solid shaft is needed to transmit 6,000 horse¬ 
power at 50 revolutions per minute if the shearing working 
strength of the material is 10,000 pounds per square inch? 






110 


STRENGTH OF MATERIALS 


We have merely to substitute in the first of the two formulas 
16 , and solve for d. Thus, 


therefore 


6,000 = 



10,000 X d 3 X 50 ^ 
321,000 

6,000 X 321,000 _ 
10,000 X 50 


3,852; 


or, d — # / 3,852 = 15.68 inches. 

(A solid shaft of this diameter contains over 25% more material than 
the hollow shaft of example 1 preceding. There is therefore considerable 
saving of material in the hollow shaft.) 

3. A solid shaft 4 inches in diameter transmits 200 horse¬ 
power while rotating at 200 revolutions per minute. What is the 
greatest shearing unit-stress in the shaft? 

We have merely to substitute in the first of the equations 16 , 
and solve for S s . Thus, 

200 - S * X 43 X 20 ° - 
— 321,000 ’ 


or. 


S = 


200 X 321,000 
4 3 X 200 


5,016 pounds per square inch. 


EXAMPLES FOR PRACTICE. 

1. What horse-power can a solid shaft 9.6 inches in diameter 
safely transmit at 50 revolutions per minute, if its shearing work¬ 
ing strength is 10,000 pounds per square inch ? 

Ans. 1,378 horse-power. 

2. What size of solid shaft is required to transmit 500 horse¬ 
power at 150 revolutions per minute, the shearing working strength 
of the material being 8,000 pounds per square inch. 

Ans. 5.1 inches. 

3. A hollow shaft whose outer diameter is 14 and inner 6.7 
inches transmits 5,000 horse-power at 60 revolutions per minute. 
What is the value of the greatest shearing unit-stress in the shaft? 

Ans. 10,273 pounds per square inch. 

STIFFNESS OF RODS, BEAMS, AND SHAFTS. 

The preceding discussions have related to the strength of 







STRENGTH OF MATERIALS 


111 


materials. We shall now consider principally the elongation of 
r °ds , deflection of beams , and twist of shafts. 

95. Coefficient of Elasticity. According to Hooke’s Law 
(Art. 9, p. 7), the elongations of a rod subjected to an increasing 
pull are proportional to the pull, provided that the stresses due to 
the pull do not exceed the elastic limit of the material. Within 
the elastic limit, then, the ratio of the pull and the elongation is 
constant; hence the ratio of the unit-stress (due to the pull) to the 
unit-elongation is also constant. This last-named ratio is called 
u coefficient of elasticity.” If E denotes this coefficient, S the 
unit-stress, and s the unit-deformation, then 


Coefficients of elasticity are usually expressed in pounds per square inch. 

The preceding remarks, definition, and formula apply also to 
a case of compression, provided that the material being compressed 
does not bend, but simply shortens in the direction of the com¬ 
pressing forces. The following table gives the average values of 
the coefficient of elasticity for various materials of construction: 


TABLE G. 

Coefficients of Elasticity. 


Material. 

Average Coefficient of Elasticity. 

Steel. 

Wrought iron. 

Cast iron. 

Timber. 

30,000,000 pounds per square inch. 
27,500,000 “ “ “ 

15,000,000 

1,800,000 “ “ “ 


The coefficients of elasticity for steel and wrought iron, for different 
grades of those materials, are remarkably constant; but for different grades 
of east iron the coefficients range from about 10,000,000 to 30,000,000 pounds 
per square inch. Naturally the coefficient has not the same value for the 
different kinds of wood; for the principal woods it ranges from 1,600,000 
(for spruce) to 2,100,000 (for white oak). 

Formula 17 can be put in a form more convenient for use, as 
follows : 

Let P denote the force producing the deformation ; A the 
area of the cross-section of the piece on which P acts ; l the length 
of the piece ; and D the deformation (elongation or shortening). 















112 


STRENGTH OF MATERIALS 


Then 

S = P -7- A (see equation 1), 
and s == D -r- l (see equation 2). 

Hence, substituting these values in equation 17, we have 


E 


P l 

AD 


; or D 


VI 

AE 


(170 


The first of these two equations is used for computing the value of 
the coefficient of elasticity from measurements of a u test,” and 
the second for computing the elongation or shortening of a given 
rod or bar for which the coefficient is known. 

Examples. 1. It is required to compute the coefficient of 
elasticity of the material the record of a test of which is given on 
pnge 9. 

Since the unit-stress S and unit-elongation s are already 
computed in that table, we can use equation 17 instead of the first 
of equations 17'. The elastic limit being between 40,000 and 
45,000 pounds per square inch, we may use any value of the 
unit-stress less than that, and the corresponding unit-elongation. 

Thus, with the first values given, 


« 


5,000 
“ 0.00017 


29,400,000. 


With the second, 


__ 10,000 

E “ 0.00035 


= 28,000,000. 


This lack of constancy in the value of E as computed from different 
loads in a test of a given material, is in part due to errors in measuring the 
deformation, a measurement difficult to make. The value of the coefficient 
adopted from such a test, is the average of all the values of E which can be 
computed from the record. 

2. Ilow much will a pull of 5,000 pounds stretch a round 
steel rod 10 feet long and 1 inch in diameter? 

We use the second of the two formulas 17'. Since A = 
0.7854 X l 2 = 0.7854 square inches, l = 120 inches, and E = 
30,000,000 pounds per square inch, the stretch is: 


D 


5,000 X 120 
0.7854 X 30,000,000 


0.0254 inch. 


















STRENGTH OF MATERIALS 


113 


EXAMPLES FOR PRACTICE. 

1. Wli at is tlie coefficient of elasticity of a material if a pull 
of 20,000 pounds will stretch a rod 1 inch in diameter and 4 feet 
long 0.045 inch ? 

Ans. 27,000,000 pounds per square inch. 

2. How much will a pull of 15,000 pounds elongate a round 
cast-iron rod 10 feet loner and 1 inch in diameter ? 

O 

Ans. 0.152 inch. 

96. Temperature Stresses. In the case of most materials, 
when a bar or rod is heated, it lengthens; and when cooled, it 
shortens if it is free to do so. The coefficient of linear expansion 
of a material is the ratio which the elongation caused in a rod or 
bar of the material by a change of one degree in temperature bears 
to the length of the rod or bar. Its values for Fahrenheit degrees 
are about as follows: 

For Steel, 0.0000065. 

For Wrought iron, .0000067. 

For Cast iron, .0000062. 

Let K be used to denote this coefficient; t a change of tem¬ 
perature, in degrees Fahrenheit; l the length of a rod or bar; 
and D the change in length due to the change of temperature. 
Then 

D = K tl. (18) 

D and l are expressed in the same unit. 

If a rod or bar is confined or restrained so that it cannot 
change its length when it is heated or cooled, then any change in 
its temperature produces a stress in the rod; such are called tem= 
perature stresses. 

j Examples. 1. A steel rod connects two solid walls and is 
screwed up so that the unit-stress in it is 10,000 pounds per 
square inch. Its temperature falls 10 degrees, and it is observed 
that the walls have not been drawn together. What is the temper¬ 
ature stress produced by the change of temperature, and what is 
the actual unit-stress in the rod at the new temperature ? 

Let l denote the length of the rod. Then the change in 
length which would occur if the rod were free, is given by formula 
18, above, thus: 

D = 0.0000005 X 10 X l = 0.000005 l. 




114 


STRENGTH OF MATERIALS 


Now, since the rod could not shorten, it has a greater than normal 
length at the new temperature; that is, the fall in temperature has 
produced an effect equivalent to an elongation in the rod amount¬ 
ing to D, and hence a tensile stress. This tensile stress can be 
computed from the elongation D by means of formula 17. Thus, 

S = Es; 


and since s , the unit-elongation, equals 


D 

/ 


.0000065 l 

l 


.0000065, 


S = 30,000,000 X .0000065 = 195.0 pounds per square inch. 
This is the value of the temperature stress; and the new unit, 
stress equals 

10,000 + 195.0 = 10,195 pounds per square inch. 

Notice that the unit temperature stresses are independent of the length 
of the rod and the area of its cross-section. 

2. Suppose that the change of temperature in the preceding 
example is a rise instead of a fall. "What are the values of the 
temperature stress due to the change, and of the new unit-stress in 
the rod ? 


The temperature stress is the same as in example 1, that is, 
1,950 pounds per square inch; but the rise in temperature 
releases, as it were, the stress in the rod due to its being screwed 
up, and the final unit stress is 

10,000 - 1,950 = 8,050 pounds per square inch. 


EXAflPLE FOR PRACTICE. 

1. The ends of a wrought-iron rod 1 inch in diameter are 
fastened to two heavy bodies which are to be drawn together, the 
temperature of the rod being 200 degrees when fastened to the ob¬ 
jects. A fall of 120 degrees is observed not to move them. 
What is the temperature stress, and what is the pull exerted by 
the rod on each object ? 

( Temperature stress, 22,000 pounds per square inch. 

AnS * \ Pull, 17,280 pounds. 

97. Deflection of Beams. Sometimes it is desirable to know 
how much a given beam will deflect under a given load, or to design 




STRENGTH OF MATERIALS 


115 


a beam which will not deflect more than a certain amount under a 
given load. In Table B, page 53, Part I, are given formulas for 
deflection in certain cases of beams and different kinds of loading. 

In those formulas, d denotes deflection; I the moment of inertia of the 
cross-section of the beam with respect to the neutral axis, as in equation 6; 
and E the coefficient of elasticity of the material of the beam (for values, see 

Art. 95). 

In each case, the load should be expressed in pounds, the length in 
inches, and the moment of inertia in biquadratic inches; then the deflection 
wall be in inches. 

According to the formulas for d , the deflection of a beam 
varies inversely as the coefficient of its material (E) and the mo¬ 
ment of inertia of its cross-section (I) ; also, in the first four and 
last two cases of the table, the deflection varies directly as the cube 
of the length (7 3 ). 

Exam/ple. What deflection is caused by a uniform load of 
6,400 pounds (including weight of the beam) in a wooden beam 
on end supports, which is 12 feet long and 6 X 12 inches in 
cross-section ? (This is the safe load for the beam ; see example 
1, Art. 65.) 

The formula for this case (see Table B, page 53) is 

5 W l 3 

d — 384 El • 

Here W = 6,400 pounds ; l = 144 inches ; E = 1,800,000 
pounds per square inch ; and 

I = Ixd = 12 6 X 12 3 = 864 inches 4 . 


Hence the deflection is 


5 X 6,400 X 144 3 
d ~~ 384 X 1,800,000 X 864 


= 0.16 inch. 


EXAMPLES FOR PRACTICE. 

1. Compute the deflection of a timber built-in cantilever 
8X8 inches which projects 8 feet from the wall and bears an 
end load of 900 pounds. (This is the safe load for the cantilever, 
see example 1, Art. 65.) 

Ans. 0.43 inch. 

2. Compute the deflection caused by a uniform load of 40,000 





116 


STRENGTH OF MATERIALS 


pounds on a 42-pound 15-inch steel I-beam which is 16 feet long 
and rests on end supports. 

Ans. 0.28 inch. 

98. Twist of Shafts. Let Fig. 57 represent a portion of a 
shaft, and suppose that the part represented lies wholly between 



two adjacent pulleys on a shaft to which twisting forces are applied 
(see Fig. 54). Imagine two radii ma and nb in the ends of the 
portion, they being parallel as showm when the shaft is not twisted. 
After the shaft is twisted they will not be parallel, ma having 
moved to ma\ and nb to nb '. The angle between the two lines in 
their twisted positions (ma' and nb ') is called the angle of twist, 
or angle of torsion, for the length l. If a'a ' is parallel to ab , then 
the angle a"nb' equals the angle of torsion. 

If the stresses in the portion of the shaft considered do not 
exceed the elastic limit, and if the twisting moment is the same 
for all sections of the portion, then the angle of torsion a (in 
degrees) can be computed from the following: 

For solid circular shafts, 


584 T^ 36,800,000 H l 

a ~ E 1 d l ~ WWn ' 
For hollow circular shafts, 

^ _ 584 T Id 36,800,000 HZ 

a ~ E , (<2‘-<Z 1 ‘) == E 1 (d* -d‘) n 




Here T, Z, d , d x , LI, and n have the same meanings as in Arts. 93 
and 94, and should be expressed in the units there used. The 
letter E 1 stands for a quantity called coefficient of elasticity for 
shear; it is analogous to the coefficient of elasticity for tension and 
compression (E), Art. 95. The values of E 1 for a few materials 
average about as follows (roughly E 1 = | E): 

















STRENGTH OF MATERIALS 


117 


For Steel, 11,000,000 pounds per square inch. 

For Wrought iron, 10,000,000 “ “ “ “ 

For Cast iron, 6,000,000 “ “ “ “ 

Example. Wliat is tlie value of the angle of torsion of a 
steel shaft 60 feet long when transmitting 6,000 horse-power at 
50 revolutions per minute, if the shaft is hollow and its outer and 
inner diameters are 16 and 8 inches respectively ? 

Here l — 720 inches; hence, substituting in the appropriate 
formula (19), we find that 

36,800,000 X 6,000 X 720 , „ J 

a ' 11,000,000 X (I6‘ - 8‘) 50 = 4 ' ‘ de S rees - 

EXAMPLE FOR PRACTICE. 

Suppose that the first two pulleys in Fig. 51 are 12 feet 
apart; that the diameter of the shaft is 2 inches; and that Pj — 100 
pounds, and a x = 15 inches. If the shaft is of wrought iron, 
what is the value of the angle of torsion for the portion between 
the first two pulleys ? 

Ans. 3.15 degrees. 

99 . Non=elastic Deformation. The preceding formulas for 
elongation, deflection, and twist hold only so long as the greatest 
unit-stress does not exceed the elastic limit. There is no theory, 
and no formula, for non-elastic deformations, those corresponding 
to stresses which exceed the elastic limit. It is well known, how¬ 
ever, that non-elastic deformations are not proportional to the 
forces producing them, but increase much faster than the loads. 
The value of the ultimate elongation of a rod or bar (that is, the 
amount of elongation at rupture), is quite well known for many 
materials. This elongation, for eight-inch specimens of various 
materials (see Art. 16), is : 

For Cast iron, about 1 per cent. 

For Wrought iron (plates), 12 - 15 per cent. 

For “ “ (bars), 20-25 “ c ‘ . 

For Structural steel, 22 - 26 “ “ . 

Specimens of ductile materials (such as wrought iron and 
structural steel"), when pulled to destruction, neck down, that is, 
diminish very considerably in cross-section at some place along 
the length of the specimen. The decrease in cross-sectional area 








118 


STRENGTH OF MATERIALS 


is known as reduction of area, and its value for wrought iron and 
steel may be as much as 50 per cent. 


RIVETED JOINTS. 

ioo. Kinds of Joints. A lap joint ia one in which the 

plates or bars joined overlap each other, as in Fig. 58, a. A butt 
joint is one in w T hich the plates or bars that are joined butt against 
each other, as in Fig. 58, b. The thin side plates on butt joints 





t> 


Fig. 58. 



are called cover=pIates; the thickness of each is always made not 
less than one-half the thickness of the main plates, that is, the 
plates or bars that are joined. Sometimes butt joints are made 
with only one cover-plate; in such a case the thickness of the 
cover-plate is made not less than that of the main plate. 

When wide bars or plates are riveted together, the rivets are 
placed in rows, always parallel to the “ seam ” and sometimes also 
perpendicular to the seam; but when we speak of a row of rivets, 
we mean a row parallel to the seam. A lap joint w T ith a single 
row of rivets is said to be single=riveted; and one w T ith two rows 
of rivets is said to be double=riveted. A butt joint w T ith two rows 
of rivets (one on each side of the joint) is called “ single-riveted,” 
and one with four rows (two on each side) is said to be ‘‘double- 
riveted.” 

The distance between the centers of consecutive holes in a 
row of rivets is called pitch. 

iok. Shearing Strength, or Shearing Value, of a Rivet. 

When a lap joint is subjected to tension (that is, when P, Fig. 58, 
is a pull), and when the joint is subjected to compression (when 
P is a push), there is a tendency to cut or shear each rivet along 
the surface between the two plates. In butt joints with two cover- 
























STRENGTH OF MATERIALS 


119 


plates, there is a tendency to cut or shear each rivet on two sur¬ 
faces (see Mg. 58, &). Therefore the rivets in the lap joint are 
said to be in single shear ; and those in the butt joint (two covers) 
are said to be in double shear. 

The u shearing value ” of a rivet means the resistance which 
it can safely offer to forces tending to shear it on its cross-section. 
This value depends on the area of the cross-section and on the work¬ 
ing strength of the material. Let d denote the diameter of the 
cross-section, and S s the shearing working strength. Then, since 
the area of the cross-section equals 0.7854 cZ 2 , the shearing strength 
of one rivet is : 

For single shear, 0.7854 d 2 S s . 

For double shear, 1.5708 d 2 S 8 . 

102. Bearing Strength, or Bearing Value, of a Plate. When 

a joint is subjected to tension or compression, each rivet presses 
against a part of the sides of the holes through wdiich it passes. 
By “ bearing value ” of a plate (in this connection) is meant the 
pressure, exerted by a rivet against the side of a hole in the plate, 
wTiich the plate can safely stand. This value depends on the 
thickness of the plate, on the diameter of the rivet, and on the 
compressive working strength of the plate. Exactly how it 
depends on these three qualities is not known; but the bearing 
value is always computed from the expression t d S c , wherein t 
denotes the thickness of the plate; d , the diameter of the rivet or 
hole; and S c , the working strength of the plate. 

103. Frictional Strength of a Joint. When a joint is sub¬ 
jected to tension or compression, there is a tendency to slippage 
between the faces of the plates of the joint. This tendency is 
overcome wholly or in part by frictional resistance between the 
plates. The frictional resistance in a well-made joint may be 
very large, for rivets are put into a joint hot, and are headed or 
capped before being cooled. In cooling they contract, drawing the 
plates of the joint tightly against each other, and producing a 
great pressure between them, which gives the joint a correspond¬ 
ingly large frictional strength. It is the opinion of some that 
all well-made joints perform their service by means of their 
frictional strength; that is to say, the rivets act only by pressing 
the plates together and are not under shearing stress, nor 


120 


STRENGTH OF MATERIALS 


are the plates under compression at the sides of their holes. The 
“ frictional strength ” of a joint, however, is usually regarded as 
uncertain, and generally no allowance is made for friction in com¬ 
putations on the strength of riveted joints. 

104. Tensile and Compressive Strength of Riveted Plates. 
The holes punched or drilled in a plate or bar weaken its tensile 
strength, and to compute that strength it is necessary to allow for 
the holes. By net section, in this connection, is meant the small¬ 
est cross-section of the plate or bar ; this is always a section along 
a line of rivet holes. 

If, as in the foregoing article, t denotes the thickness of the 
plates joined ; d , the diameter of the holes; n v the number of riv¬ 
ets in a row ; and w, the width of the plate or bar; then the net 
section = (w - n/1) t. 

Let S t denote the tensile working strength of the plate ; then 
the strength of the unriveted plate is wtS„ and the reduced tensile 
strength is (w - n l d') t S t . 

The compressive strength of a plate is also lessened by the 
presence of holes ; but when they are again filled up, as in a joint, 
the metal is replaced, as it were, and the compressive strength of 
the plate is restored. No allowance is therefore made for holes in 
figuring the compressive strength of a plate. 

105. Computation of the Strength of a Joint. The strength 
of a joint is determined by either (1) the shearing value of the 
rivets ; (2) the bearing value of the plate ; or (3) the tensile 
strength of the riveted plate if the joint is in tension. Let P 9 de¬ 
note the strength of the joint as computed from the shearing 
values of the rivets ; P c , that computed from the bearing value of 
the plates ; and P t , the tensile strength of the riveted plates. 
Then, as before explained, 

p t= { w - n i d ) *s»; ) 

P s = w 2 0.7854 rf 2 S,; and V ( 20 ) 

P c = njd S c ; ) 

denoting the total number of rivets in the joint ; and n 3 denot¬ 
ing the total number of rivets in a lap joint, and one-half the 
number of rivets in a butt joint. 

Examples. 1. Two half-inch plates 7^ inches wide are con- 


STRENGTH OF MATERIALS 


121 


nected by a single lap joint double-riveted, six rivets in two rows. 
If the diameter of the rivets is & inch, and the working strengths 
are as follows : S t ~ 12,000, S 9 = 7,500, and S c = 15,000 pounds 
per square inch, what is the safe tension which the joint can 
transmit ? 

Here n== 3, n== 6, and n== 6 ; hence 

1 3 1 

P t = (7-g— 3 X -j-) X -g- X 12,000 = 31,500 pounds; 

P s = 6 X 3.7854 X (A) J X 7,500 = 19,880 pounds ; 

1 3 

P c = 6 X -o" X ~r X 15,000 = 33,750 pounds. 

Since P s is the least of these three values, the strength of the 
joint depends on the shearing value of its rivets, and it equals 
19,880 pounds. 

2. Suppose that the plates described in the preceding example 
are joined by means of a butt joint (two cover-plates), and 12 
rivets are used, being spaced as before. What is the safe tension 
which the joint can bear ? 

Here n x = 3, n 2 = 12, and n z = 6; hence, as in the preced¬ 
ing example, 

P t = 31,500; and P c = 33,750 pounds; but 

P s = 12 X 0.7854 X (A) 2 x 7,500 = 39,760 pounda 

The strength equals 31,500 pounds, and the joint is stronger than 
the first. 

3. Suppose that in the preceding example the rivets are 
arranged in rows of two. What is the tensile strength of th& 
joint ? 

Here n x = 2. n 2 = 12, and n 3 — 6; hence, as in the preced¬ 
ing example, 

P s = 39,760; and P c = 33,750 pounds; but 

P t = (7 A_2 X A) A X 12,000 = 36,000 pounds. 

The strength equals 33,750 pounds, and this joint is stronger than 
either of the first two. 












122 


STRENGTH OF MATERIALS 


EXAMPLES FOR PRACTICE. 

Note. Use working strengths as in example 1, above. 

St = 12,000, S s = 7,500, and S c = 15,000 pounds per square inch. 

1. Two half-inch plates 5 inches wide are connected by a 
lap joint, w r ith two |-inch rivets in a row. What is the safe 
strength of the joint ? 

Ans. 6,625 pounds. 

2. Solve the preceding example supposing that four J-inch 
rivets are used, in two rows. 

Ans. 13,250 pounds. 

3. Solve example 1 supposing that three 1-inch rivets are 

used, placed in a row lengthwdse of the joint. 

Ans. 17,670 pounds. 

4. Two half-inch plates 5 inches wide are connected by a 
butt joint (two cover-plates), and four |-inch rivets are used, in 
two rows. What is the strength of the joint ? 

Ans. 11,250 pounds. 

106. Efficiency of a Joint. The ratio of the strength of a 
joint to that of the solid plate is called the “ efficiency of the 
joint.” If ultimate strengths are used in computing the ratio, 
then the efficiency is called ultimate efficiency; and. if working 
strengths are used, then it is called working efficiency. In the 
following, w r e refer to the latter. An efficiency is sometimes ex¬ 
pressed as a per cent. To express it thus, multiply the ratio 
strength of joint -r- strength of solid jplcite, by 100. 

Example . It is required to compute the efficiencies of the 
joints described in the examples worked out in the preceding article. 

In each case the plate is 4 inch thick and 7-| inches wide; 
hence the tensile working strength of the solid plate is 

7-y- X 4 x 12,000 = 45,000 pounds. 


Therefore the efficiencies of the joints are : 
19,880 ... .. 

(1) iKnnn = °' 44 ’ or 44 P er cent '’ 


( 2 ) 

( 3 ) 


45,000 

31,500 

45,000 

33,750 

4p00 


= 0.70, or 70 per cent; 
= 0.75, or 75 per cent. 






INDEX 


Page 

Angle of torsion. 116 

Angle of twist. *116 

Applications of first beam formula. 59 

Beam formula 

first. 59 

second. 73 

Beams. 19 

deflection of. 114 

stiffness of. 110 

Bearing strength of a plate. 119 

Bending moment.23, 32 

maximum. 40 

Brick, ultimate strength of. 15 

Broken straight-line formula. 101 

Built-up sections, moment of inertia of. 49 

Butt joint. 118 

Cast iron 

intension. 13 

in compression. 15 

Center of gravity of an area. 41 

Center of gravity of built-up sections. 44 

Centrally loaded column. 88 

Coefficient of elasticity.„. Ill 

for shear. 116 

Coefficient of linear expansion. 113 

Column formulas 

graphical representation of. 93 

Rankine’s. 88 

Column loads, kinds of. 88 

Columns 

classes of. 86 

cross-sections of. 86 

design of. 102 

end conditions of. 85 

strength of. 85 

Combination column formulas. 94 

Combined flexural and direct stress. 83 

Compression, materials in. 13 

Compressive strength of riveted plates. 120 

Continuous beam.. 19 








































124 


INDEX 


Page 

Cover-plates. 118 

Deflection of beams. 114 

Deformation. 4 

Design of 

columns. 102 

timber beams. 76 

Direct stress. 79 

Double shear. 119 

Eccentrically loaded. 88 

Efficiency of a joint. 122 

Elastic limit. 5 

Elasticity. 4 

End conditions of column. 85 

Euler formulas. 94 

External shear. 23 

Factor of safety.'. 8 

Fibre stresses.55, 57 

First beam formula. 59 

Flexural stress. 79 

Flexure and compression. 81 

Flexure and tension. 79 

Formula for power which shaft can transmit. 109 

Formula for strength of shaft. 107 

Frictional strength of joint. 119 

Graphical representation of column formulas. 93 

Hooke’s law. 5 

Horizontal shear. 75 

Inclined forces. 54 

Joint 

efficiency of. 122 

frictional strength of. 119 

Kinds of loads and beams.52, 79 

Kinds of stress. 2, 55 

Lap joint. 118 

Laws of strength of beams. 71 

Linear expansion, coefficient of. 113 

Longitudinal forces. 54 

Main plates. 118 

Materials 

in compression. 13 

in shear. 15 

in tension. 11 

Maximum bending moment. 40 

Maximum shear. 31 

Metals in shear. 16 

Modulus of rupture. 72 

Moment diagrams. 36 

Moment of a force. 16 















































INDEX 125 

Page 

Moment of inertia... 40 

of built-up sections. 49 

of a rectangle. 40 

unit of. 47 

Moments, principle of. 17 

Neutral axis. 54 

Neutral line. 04 

Neutral surface. 54 

Non-elastic deformation. 117 

Notation. 24 33 

Parabola-Euler formulas. pg 

Posts . 85 

Principle of moments. yj 

applied to areas. 42 

Radius of gyration. gp 

Rankine’s column formula. gg 

Reactions of supports. 15 

Rectangle, moment of inertia of. 4 g 

Reduction formula. 4 g 

Resisting moment. 57 ^ 107 

value of. 5 g 

Resisting shear. 73 

Restrained beam. 19 

Rivet, shearing strength of. 118 

Riveted joints. Hg 

Rods, stiffness of. 110 

Rule of signs.23, 32 

Safe load of a beam. 71 

Safe strength of a beam. 71 

Second beam formula. 73 

Section modulus. 59 

Shafts 

stiffness of. 110 

strength of. 105 

twist of. 116 

twisting moment of. 105 

Shear diagrams. 27 

Shear stress. 2 

Shearing strength of rivet. 118 

Shears, unit for. 24 

Signs, rule of. 23, 32 

Simple beam.*. 19 

Simple stress. 1 

Single-shear. 119 

Steel in compression. 14 

Steel in tension. 12 

Stiffness of rods, beams, and shafts. 110 

Stone, ultimate strength of. 15 

















































126 


INDEX 


Page 

Straight-line formulas. 94 

Strength of beams. 52 

laws of. 71 

Strength of columns. 85 

Strength of joint, computation of. 120 

Strength of shafts. 105 

Stress, kinds of. 2 

Stress-deformation diagram. 0 

Struts. 85 

Tables 

bending moment. 53 

coefficients of elasticity. Ill 

deflection. 53 

factors of safety. 10 

maximum shear, values of. 53 

mild steel columns, data for.95, 99 

moduli of rupture. 72 

moment diagrams. 53 

moments of inertia. 52 

properties of standard I-beams. 70 

radii of gyration. 52 

section moduli. 52 

shear diagrams. 53 

Temperature stresses. 113 

Tensile strength of riveted plates. 120 

Tension, materials in. 11 

Timber 

in compression. 14 

in shear. 15 

in tension. 11 

Timber beams, design of. 76 

Torsional stress. 106 

Transverse forces. 52 

Twist of shafts. 116 

Twisting moment of shaft. 105 

Ultimate strength. 6 

Unit-deformation. 4 

Unit of moment of inertia. 47 

Unit-stress. 3 

Units. 33 

Units for shears. 24 

Value of resisting moment. 58 

Wrought-iron 

in compression. 14 

intension. 12 

Working strength. 8 

Working stress. 8 















































( 






41 




/ 






J 




























































































































































